Unit 1 - Kinematics

1. Kinematics

Title

AP Physics C: Mechanics — Unit 1

Position, velocity and acceleration as derivatives and integrals — the language every other unit is written in

2. What this deck buys you

Objectives

Unit 1 is not really about motion. It is about learning to say where something is as a function of time, and then letting calculus do the rest.

• Move between position, velocity and acceleration by differentiating and integrating
• Derive the constant-acceleration equations instead of memorising four unrelated formulas
• Read position, velocity and acceleration graphs as slopes and areas
• Split two-dimensional motion into two independent one-dimensional problems
• Handle acceleration that is not constant — the thing that separates Physics C from Physics 1

The AP Physics 1 version of this unit hands you formulas. The C version hands you a derivative. Everything below is the same physics with the calculus put back in — which is why the hard free-response problems become easier, not harder.

If you have already taken a physics course, the first half of this deck will look familiar and the second half will not. That is deliberate, and it is worth knowing which is which before you start.

The AP Physics 1 treatment of motion gives you four equations and a lot of practice choosing between them. It works, but it hides where those equations came from, and it quietly assumes the acceleration never changes. Physics C removes that assumption, and the price of removing it is calculus.

So the through-line here is a single sentence: velocity is the derivative of position, acceleration is the derivative of velocity, and integration runs the whole thing backwards. Everything else in this deck — the four equations, the graphs, projectiles, relative motion — is that one sentence applied in a particular situation.

One promise about the difficulty. The calculus makes the easy problems slightly longer and the hard problems dramatically shorter. A question about a rocket whose thrust varies with time is essentially unanswerable with the Physics 1 toolkit and is a routine integration with this one.

3. Position and its Derivatives

Section

Section 1

4. Before any formula: an origin, an axis, and a positive direction

Concept

A position is not a property of the object. It is a statement about the object and your coordinate system together. Change the origin and every position changes; change which way is positive and every sign flips.

This sounds like bookkeeping. It is the single most common source of lost points on free-response kinematics, because a sign convention chosen halfway through a problem is a sign convention that will be forgotten.

Write the axis on your diagram before you write anything else. Then never change it for the rest of the problem, even when the object turns around.

This slide looks like preamble and it is not. It is the step that decides whether the rest of your work will have correct signs, and it is the step most people skip when they are in a hurry.

Here is the thing to internalise: a position is not something an object has by itself. It is a statement about the object and about a coordinate system that you invented. Move the origin two metres left and every position in the problem changes by two metres. Decide that left is positive instead of right and every sign in the problem flips.

None of that is a problem — right up until you make the choice twice. The classic disaster is a ball thrown upward where you take up as positive at the start, and then, when the ball begins falling, you start thinking of the downward motion as positive because it feels natural. Now half your equation is in one coordinate system and half in another, and the answer is wrong in a way that is very hard to find.

So: draw the axis, mark the positive direction with an arrow, and write down the origin. Then treat that as fixed for the entire problem, including the parts where the object is moving the other way. A negative answer is not a mistake to be corrected — it is your coordinate system telling you which direction the result points.

5. Reading a displacement: why it is not a distance

Notation

Distance is the length of the path travelled and is never negative. Displacement is the change in position and carries a sign. On a round trip the distance is large and the displacement is exactly zero.

Every quantity in this unit comes in a pair like this: distance and displacement, speed and velocity. The scalar member of each pair adds up the whole journey; the vector member only compares the endpoints.

Distance and displacement get introduced together in every physics course, everyone nods, and then half the class loses points on a problem where the difference matters. It is worth being precise about what separates them.

Displacement only knows two things: where you started and where you finished. It is deliberately, aggressively ignorant of the route. Distance is the opposite — it is the odometer reading, the total length of path actually covered, and it never decreases.

The cleanest test case is a round trip. Drive to the shop and back and your distance is, say, six kilometres while your displacement is exactly zero. Neither number is wrong; they answer different questions. If someone asks how much fuel you used, they want the distance. If they ask where you are now, they want the displacement.

Watch the order in the delta as well. Final minus initial, always, in that order. Writing initial minus final is not an alternative convention you can adopt consistently — it is a sign error, because every other formula in the course assumes the standard order.

6. Average velocity is the slope of a secant line

Concept

\[ v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i} \]

Average velocity answers a narrow question: if the motion had been steady, how fast would it have had to be to cover this displacement in this time? It says nothing about what actually happened in between.

Notice that the graph and the formula are the same statement. Rise over run is change in position over change in time. If you can see a slope, you can see an average velocity.

The formula and the graph on this slide are the same statement written two ways, and being able to switch between them freely is most of what graph questions test.

Algebraically, average velocity is change in position over change in time. Graphically, that is rise over run between two points on the position graph — which is exactly the definition of the slope of the straight line joining them. A secant line, in the language of your calculus course.

Be honest about what the average velocity actually tells you, though, because it is less than people assume. It answers: if this motion had been perfectly steady, what speed would have been needed to cover this displacement in this time? A car that drives fast, stops for lunch, and drives fast again has the same average velocity as one that cruised gently the whole way. The average has thrown away everything that happened in between.

This is also why average velocity is not, in general, the average of the starting and ending speeds. That shortcut works for constant acceleration and nowhere else, and there is a slide later in this deck about exactly why.

7. Instantaneous velocity is what the secant becomes

Concept

\[ v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} \]

This is the whole idea of the C course in one line. Velocity is not a separate quantity you look up — it is the derivative of the position function, and the tangent line is what that looks like on a graph.

So a question like how fast is it going at t equals 3 seconds is a calculus question. Differentiate, then substitute. In that order — substituting first collapses the function to a number, and the derivative of a number is zero.

This is the moment the course becomes Physics C rather than Physics 1, so it is worth slowing down for.

Look at what happens in the figure. Fix one point on the curve and slide the second point toward it. Each secant line is the average velocity over a shorter and shorter interval. The lines do not wander — they close in on one particular line, the tangent at that point, and the slope of that line is the instantaneous velocity.

That limiting process is precisely the definition of a derivative, which means velocity is not a new quantity you have to look up in a table. It is dx/dt, full stop. If you can differentiate, you can find the velocity of anything whose position you can write down.

One procedural warning that costs marks constantly. If a question asks for the velocity at t equals three seconds, differentiate first and substitute second. Substituting first turns your position function into a single number, and the derivative of a number is zero — which is how people end up confidently reporting that a moving object has zero velocity.

8. The tangent line is the speedometer

Picture it

Here is the mental picture worth keeping. The position graph is the road; your speedometer reads the steepness of the road at the point you are standing on.

Three readings you should be able to take off any position graph without computing anything: steep means fast, flat means momentarily at rest, downhill means negative velocity.

A flat spot is not the object resting for a while — it is one instant at which the velocity passes through zero. To rest for a while the graph would have to be flat over an interval.

If you take one picture away from this section, make it this one. The position graph is a road seen from the side, and the velocity is how steep the road is at the point where you are standing.

Three readings you should be able to make instantly, without calculating anything. A steep tangent means fast. A flat tangent means momentarily at rest. A tangent sloping downward means the object is moving in the negative direction — not slowing down, but actually travelling backwards along your axis.

The middle one deserves care. A single flat point is not the object taking a break; it is one instant at which the velocity happens to pass through zero, usually on its way from positive to negative. For an object to genuinely sit still for a while, the graph would have to be flat over a whole interval, not at a single point.

Practise reading graphs this way before you practise computing from them. On the exam, a great many graph questions can be answered by looking at three tangents and never picking up a pencil.

9. From x(t) to v(t) to a(t)

Worked example

A particle moves along a straight line with position given in metres, time in seconds.

\[ x(t) = 2t^3 - 9t^2 + 12t + 5 \]

\[ v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12 \]

\[ a(t) = \frac{dv}{dt} = 12t - 18 \]

\[ 6t^2 - 18t + 12 = 6(t-1)(t-2) = 0 \quad \Rightarrow \quad t = 1\,\text{s},\; t = 2\,\text{s} \]

Read the story back: the particle moves forward, stops at one second, reverses, stops again at two seconds, then moves forward for good. Two turning points, from one cubic.

This is the basic exercise of the unit, and every harder problem contains it somewhere. Given a position function, differentiate twice and read off the story.

The mechanical part is just the power rule. The interpretive part is where the marks are. Look at what the velocity function tells you here: it factors into six times t minus one times t minus two, so it is zero at one second and at two seconds. Those are the instants when the particle is momentarily at rest.

Now be careful with what those instants mean. The particle is at rest but not staying at rest, and you can prove that from the acceleration, which is minus six and plus six at those two times — non-zero in both cases. Something at rest with a non-zero acceleration will not be at rest an instant later. These are turning points, not stops.

Read the whole motion back as a narrative and you will remember it far better than the algebra: the particle moves forward, slows, stops at one second, reverses, moves backward, stops again at two seconds, then goes forward and keeps going. All of that from one cubic and two derivatives.

10. Acceleration is the derivative of the derivative

Concept

\[ a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} \]

Almost every kinematics problem in this course is a request to move up or down this ladder. Recognising which direction you are being asked to go is most of the work.

Going down is free — differentiate and you are done. Going up costs you a constant of integration each time, and that constant has to be paid for with an initial condition. Two rungs up means two constants and two initial conditions.

The ladder in the figure is the organising picture for the entire unit, and it is worth being able to draw it from memory.

Going down — differentiating — is free. You have a function, you apply the rules, you are done. There is no extra information required and no ambiguity in the answer.

Going up — integrating — costs you something every time, and that something is an initial condition. The integral tells you how much the quantity changed, but it cannot possibly know where it started, because that information is not contained in the rate of change. Two rungs up means two constants and therefore two initial conditions, which is why free-response problems that hand you an acceleration always also hand you an initial velocity and an initial position.

When you sit down to a kinematics problem, the very first question to ask is which direction on this ladder you are being asked to travel. Down means differentiate. Up means integrate and pay attention to the constants. Almost everything else is detail.

11. Scalar or vector?

Sorting

The distinction decides whether a minus sign is allowed to appear. Sort each quantity.

A speedometer is the honest test case. It reads 60 whether you are driving north or south, and it never reads a negative number — that is what makes speed a scalar.

The reason to be careful here is entirely practical: the answer decides whether a minus sign is allowed to appear in your working.

A scalar answers how much and nothing else. Distance, speed, elapsed time — all of them accumulate as the motion goes on and none of them can be negative. A vector answers how much and which way. In one dimension the which-way collapses into a plus or a minus sign, which is exactly why position, displacement, velocity and acceleration can all come out negative.

The speedometer is the honest test. It reads sixty whether you are driving north or south, and it has never in its life shown a negative number. That is speed being a scalar. Your velocity, meanwhile, is genuinely different in those two cases, and if you set up an axis pointing north then one of them is negative.

One consequence to watch for: because speed is the magnitude of velocity, a question asking for a speed should never get a negative answer. If yours is negative, either you meant velocity, or something upstream went wrong.

12. A negative acceleration means the object is slowing down

Trap

13. A ball thrown straight up: what happens at the very top?

Prediction

The apex is the cleanest example in the whole course of velocity and acceleration being genuinely independent. One is zero; the other is at full strength.

The apex is the single best example in the course of velocity and acceleration being genuinely independent quantities, which is why it appears on the exam so often.

At the top, the velocity really is zero. It has to be — the ball is switching from moving upward to moving downward, and the only way to get from a positive number to a negative one continuously is to pass through zero. That instant is the apex.

The acceleration, though, is at full strength. Gravity did not pause; nothing about the ball being momentarily motionless changes the fact that the Earth is pulling on it. The acceleration is 9.8 metres per second squared downward at the apex, exactly as it was on the way up and will be on the way down.

Here is the argument that makes it stick, and it generalises to a lot of physics. Suppose the acceleration were zero at the apex. Then the velocity would not be changing. But the velocity is zero, so it would stay zero — and the ball would hang motionless in mid-air indefinitely. Since balls do not do that, the acceleration is not zero.

14. Running the Derivatives Backwards

Section

Section 2

15. Integration climbs back up the ladder

Concept

\[ v(t) = v_0 + \int_0^{t} a(t')\,dt' \qquad x(t) = x_0 + \int_0^{t} v(t')\,dt' \]

Read the structure, not the symbols. Each line says: start where you started, then add up all the change that happened since. The integral is the accumulated change; the initial value is where the accumulating began.

The two constants, v0 and x0, are not decoration. They are the physics the calculus cannot supply: the integral knows how the motion changed, and only the initial condition knows where it was.

Read the structure of these two equations rather than the symbols, because the structure is the same in both and it will reappear all year.

Each says: you end up at your starting value plus everything that accumulated since. The integral is the accumulation; the initial value is where the accumulating began. Nothing more complicated than that is happening.

The figure makes the accumulation visible. Take an acceleration graph, find the area under it, and that area is the change in velocity. Not the velocity — the change. To get the velocity itself, you have to add the value you started with, which is exactly what v-nought is doing sitting outside the integral.

This is worth belabouring because it is where students lose the constant. The integral genuinely does not know your initial velocity. It cannot know. That information has to come from the problem statement, and if the problem gave it to you and you did not use it, something is missing from your answer.

16. The area under a v-t graph is the displacement

Concept

\[ \Delta x = \int_{t_i}^{t_f} v(t)\,dt \]

Because area below the axis is negative, this integral gives displacement, not distance. It already knows about the object turning around and coming back.

If a problem asks for total distance, you must integrate the absolute value of the velocity — which in practice means splitting the integral at every instant where v crosses zero and adding the pieces as positive numbers.

Areas on a velocity graph are signed, and that single fact is what makes this idea useful rather than a curiosity.

Area above the time axis counts as positive; area below counts as negative. So when you integrate a velocity function that changes sign, the integral automatically handles the object turning around and coming back. What comes out is the displacement — net change in position — with no extra bookkeeping from you.

That is also its limitation. If a question asks for the total distance travelled, the plain integral will undercount, because the backtracking cancelled some of the forward motion. For distance you need the integral of the absolute value of the velocity, which in practice means finding every time the velocity crosses zero, splitting the integral there, and adding the pieces as positive numbers.

A reliable exam habit: whenever you see the words total distance rather than displacement, immediately look for where the velocity changes sign. The question is almost certainly testing exactly that.

17. Given a(t) = 6t, recover v(t) and x(t)

Worked example

A particle starts at x equals 2 m with velocity 4 m/s, and its acceleration is a(t) = 6t.

\[ v(t) = \int 6t\,dt = 3t^2 + C_1 \]

\[ v(0) = 3(0)^2 + C_1 = 4 \quad \Rightarrow \quad C_1 = 4, \qquad v(t) = 3t^2 + 4 \]

\[ x(t) = \int (3t^2 + 4)\,dt = t^3 + 4t + C_2 \]

\[ x(t) = t^3 + 4t + 2 \]

This is the template for every integrate-upward problem, so it is worth learning the rhythm rather than the specific numbers.

The rhythm is: integrate, apply an initial condition, integrate, apply the other initial condition. Four steps, strictly alternating. What you must not do is integrate twice and then try to sort the constants out at the end.

Notice how the constant gets pinned. After the first integration you have three t-squared plus C-one. Substituting t equals zero annihilates the t-squared term and leaves C-one standing alone, equal to whatever the initial velocity was. That is why initial conditions are always given at t equals zero if the problem can arrange it — it makes the evaluation trivial.

And notice the verification step at the end. Differentiating back down to check is cheap, it catches sign and coefficient errors immediately, and on a free-response question it costs you about fifteen seconds. Get in the habit now, while the functions are simple.

18. The constant of integration is a formality you can add at the end

Trap

19. Find the mistake in this integration

Error analysis

A student is asked: a particle has a(t) = 4t with v(0) = 3 and x(0) = 1. Find x(t). Here is the work handed in.

The corrected work: v(t) = 2t squared + 3, then x(t) = (2/3)t cubed + 3t + 1.

\[ x(t) = \tfrac{2}{3}t^3 + 3t + 1 \]

The lesson is structural rather than arithmetic. Never stack two integral signs in kinematics. Integrate once, apply the initial condition, and only then integrate again.

The arithmetic in this student's work is flawless. Every antiderivative is correct. The answer is still wrong, and that combination is worth studying because it is how most real errors look.

The problem is structural. Stacking two integral signs and evaluating them in one go skips the moment where the initial velocity was supposed to enter. So the first integration produced two t squared when it should have produced two t squared plus three, and that missing three was then integrated into a missing three-t in the position.

Once that term is gone it cannot be recovered later. No amount of careful work in the second integration will put it back, and the single C at the end cannot encode two independent pieces of information. One constant cannot do the job of two.

So the rule to take away is a rule about layout rather than about calculus: never write two integral signs next to each other in a kinematics problem. Integrate once. Stop. Apply the initial condition. Then integrate again. The discipline of stopping is the entire safeguard.

20. Constant Acceleration: the Big Four

Section

Section 3

21. Where the kinematic equations actually come from

Concept

These are not four independent laws. They are what the integrals above collapse to in the one special case where the acceleration does not change.

\[ v(t) = v_0 + \int_0^t a\,dt' = v_0 + at \]

\[ x(t) = x_0 + \int_0^t (v_0 + at')\,dt' = x_0 + v_0 t + \tfrac{1}{2}at^2 \]

That is two of the four, derived in two lines. The other two follow by eliminating a variable algebraically rather than by any new physics.

This is worth doing once by hand, because it tells you exactly when the equations are allowed: only when a is constant. Every time you reach for one of them, you are asserting that. If the acceleration varies, they are simply the wrong tool and no amount of care with the algebra will save the answer.

You have probably met these four equations before as things to memorise. Deriving them once changes your relationship with them permanently, and it takes about two lines.

Start with acceleration constant. Integrating a constant with respect to time gives that constant times t, so the velocity is the initial velocity plus a-t. Integrate that and you get the initial position plus v-nought t plus one half a t squared. Two of the four equations, from two integrations, with nothing memorised.

The pay-off is not saving memory. It is knowing the condition. Look at where the derivation used the fact that a was constant: it happened at the very first integration, when a came straight out of the integral sign. If a varies with time, that step is illegal, and everything downstream of it is void.

So every time you write down one of these equations, you are making a claim — that the acceleration is constant over the whole interval you are applying it to. Most kinematics errors in this course are not arithmetic. They are applying these four equations to a situation where that claim is false.

22. Reading the big four

Notation

\[ v^2 = v_0^2 + 2a(x - x_0) \qquad x = x_0 + \tfrac{1}{2}(v_0 + v)t \]

The third equation has no t in it, and the fourth has no a. That is their entire reason for existing: each one lets you skip a quantity you neither know nor want.

All four are one-dimensional statements. In two dimensions you apply the whole set twice — once to x, once to y — with the same t linking them.

Three details in the notation cause more trouble than the equations themselves.

The subscript zero means at the instant you started the clock. It does not mean at the ground, or at the origin, or at the beginning of time. If you restart your clock partway through a two-phase problem — and you often should — then the zero-subscripted quantities are the values at that restart, not at the original launch.

The t in these equations is elapsed time, never a clock reading. A problem describing what happens between t equals two seconds and t equals five seconds has t equal to three in the equations. Substituting five is a very common and very quiet error.

And the a is assumed constant across the entire interval. A rocket that burns for four seconds and then coasts has two intervals with two different accelerations, and you must apply the equations twice rather than once.

Finally, remember these are one-dimensional statements. In two dimensions you use the whole set twice — once for x, once for y — with the same t threading through both.

23. The fourth equation, and why it has no t in it

Concept

You could get it by solving the first equation for t and substituting. There is a cleaner route that will matter again later in the course.

\[ a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx} \]

\[ \int_{x_0}^{x} a\,dx' = \int_{v_0}^{v} v'\,dv' \quad \Rightarrow \quad a(x - x_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \]

\[ v^2 = v_0^2 + 2a(x - x_0) \]

Two things to take away. First, v dv/dx is how you handle any problem where acceleration depends on position rather than time — you will meet it again in the last section of this deck and again in oscillations.

Second, look at the shape of that result. Multiply through by one half of the mass and it is the work-energy theorem. The energy chapter is already hiding inside kinematics.

You could derive this equation by solving the first one for t and substituting into the second. That works, and it is what most textbooks do. The route shown here is better, because the manoeuvre it uses will come back several times this year.

The trick is the chain rule applied to acceleration. Write a as dv/dt, then insert dx over dx and regroup, and you get a equals v times dv/dx. That expression contains no time at all — it relates acceleration to how the velocity changes with position.

Separate and integrate and you land on v squared equals v-nought squared plus two a delta x. Notice what happened: time never appeared anywhere in the derivation, which is precisely why it does not appear in the result.

Now look at the shape of that final line. Multiply through by one half of the mass and the right side becomes a change in kinetic energy while the left becomes force times distance. You have just derived the work-energy theorem, three units early, without meaning to. Energy methods and this identity are the same mathematics wearing different clothes, which is why they always agree.

24. Which equation, and why

Comparison

Fill in what each equation leaves out. The missing variable is the reason you would choose it.

Every constant-acceleration problem involves five quantities. A problem will hand you three and ask for a fourth — which means exactly one of these equations avoids the fifth. Finding it is the whole selection process.

The selection process for these four equations sounds like it should require judgement, and it does not. It is mechanical, and this table is why.

There are five quantities in play: initial position, initial velocity, final velocity, acceleration, and time. Each of the four equations contains exactly four of them and omits one. That omission is the equation's entire personality.

So a typical problem hands you three quantities and asks for a fourth. The fifth — the one nobody mentioned and nobody wants — identifies your equation uniquely, because exactly one of the four leaves it out.

Practise the identification rather than the algebra. Read the problem, list the five, mark what you have, mark what you want, and see what is left over. On a timed exam this is worth more than being fast at rearranging equations, because it eliminates the flailing.

25. The five-variable recipe

Pattern

This procedure turns almost every constant-acceleration problem into bookkeeping.

1. Draw the axis and mark the positive direction. Commit to it.
2. List the five: x0, v0, v, a, t. Write down the three you were given, with signs.
3. Circle the one you want and cross out the one nobody mentions.
4. Pick the equation that is missing the crossed-out variable. There will be exactly one.
5. Solve symbolically, then substitute numbers with units.
6. Check the sign and the size against the picture before writing the answer down.

The fifth step matters more than it looks. Substituting numbers early buries sign errors inside arithmetic, where they are almost impossible to find.

If two phases of the motion have different accelerations — a car that speeds up and then brakes — run the recipe twice. The final velocity of phase one becomes the initial velocity of phase two, and the clock restarts.

This is the procedure to run when a constant-acceleration problem appears and you are not sure where to start. It converts thinking into bookkeeping, which is what you want under time pressure.

The step people skip is the first one, drawing the axis. Do not skip it. Almost every sign error in this unit traces back to a coordinate system that was never written down and therefore drifted partway through the problem.

The step people underrate is solving symbolically before substituting. Numbers hide errors. If you substitute early, a sign mistake becomes just another digit in a long arithmetic expression and you will never find it. If you solve for the symbol first, you can check the form of the answer against your intuition before any arithmetic happens.

And when the motion has two phases with different accelerations, run the whole recipe twice. Phase one's final velocity becomes phase two's initial velocity, the clock resets to zero, and the axis — importantly — stays exactly as it was.

26. Pick the equation

Check

A sled starts from rest and accelerates uniformly down a slope, covering 40 m before reaching a speed of 20 m/s.

Work this one by the recipe rather than by recognition, because the recipe is what will still work when the numbers are unfamiliar.

List the five. You are given the initial velocity, which is zero because it starts from rest. You are given the final velocity, twenty. You are given the displacement, forty. You are asked for the acceleration. That leaves time as the quantity that is neither given nor wanted.

Exactly one equation omits time, and that is the one to use. Four hundred equals zero plus two a times forty, so a is five metres per second squared. One line, no intermediate quantities.

It is worth noticing what the other three options cost. Every one of them contains t, so choosing any of them turns a one-step problem into a two-step problem where you first solve for a time you were never asked about. That is not wrong, just slower — and on an exam, slower is a kind of wrong.

27. A car brakes to a stop

Worked example

A car travelling at 28 m/s brakes with a constant acceleration of magnitude 7 m/s squared. How far does it travel before stopping, and how long does it take?

\[ v^2 = v_0^2 + 2a\Delta x \;\Rightarrow\; 0 = 28^2 + 2(-7)\Delta x \;\Rightarrow\; \Delta x = \frac{784}{14} = 56\,\text{m} \]

\[ 0 = 28 + (-7)t \quad \Rightarrow \quad t = 4\,\text{s} \]

Worth noticing: stopping distance goes as the square of the speed. Double the speed and the stopping distance quadruples — this is the physics behind every highway safety campaign you have ever seen.

The setup step matters more than the algebra here. Braking means the acceleration opposes the motion, so once you have chosen the direction of travel as positive, the acceleration must carry a minus sign. Getting that sign in is the whole problem; everything after it is substitution.

For the distance, time is neither given nor wanted, so the no-time equation is the direct route. For the time, displacement has already been found but is not needed, so the linear equation is simpler. Two different questions about the same motion, two different equations, chosen by the same criterion each time.

The verification with the fourth equation is worth doing. Average velocity for constant acceleration is the plain average of the endpoints, fourteen metres per second, and fourteen times four seconds is fifty-six metres. Two independent routes agreeing is strong evidence that both the algebra and the signs are right.

The physical result deserves a moment too. Stopping distance depends on the square of the speed. Going from fifty to a hundred kilometres per hour does not double your stopping distance, it quadruples it. That single quadratic is behind essentially every road safety campaign you have ever seen.

28. Two phases: powered climb, then coast

Worked example

A model rocket accelerates upward from rest at 30 m/s squared for 4.0 s, then the engine cuts out. Find the maximum height. Take g = 9.8 m/s squared.

\[ y_1 = \tfrac{1}{2}(30)(4)^2 = 240\,\text{m}, \qquad v_1 = (30)(4) = 120\,\text{m/s} \]

\[ 0 = (120)^2 + 2(-9.8)\Delta y \;\Rightarrow\; \Delta y = \frac{14400}{19.6} \approx 735\,\text{m} \]

\[ h \approx 9.8 \times 10^{2}\,\text{m} \]

The structural point: the engine cut-out is a boundary, and the kinematic equations may not be applied across it. Phase one's ending is phase two's beginning, and the clock resets.

Two-phase problems are where most people first meet the limits of the big four, and the structural lesson is more important than the numbers.

The engine cut-out is a boundary. Before it the acceleration is plus thirty; after it the acceleration is minus nine point eight. There is no single constant acceleration that describes the whole flight, so there is no single application of the kinematic equations that can cover it. Trying to average the two accelerations is a tempting shortcut and it is simply wrong.

The correct structure is a handoff. Solve phase one completely, getting both the height and the velocity at burnout. Those two numbers become the initial conditions for phase two, the clock restarts at zero, and you run the equations again.

Keep the axis fixed across the boundary. Up was positive during the powered climb and up is still positive during the coast, which is why the acceleration changes sign — the physics changed, not the coordinate system.

Once you can see the boundary, these problems stop being harder than single-phase ones. They are just two of them stapled together, and recognising where to staple is the skill being tested.

29. Why does the average-velocity shortcut need constant a?

Explain it to yourself

For constant acceleration, the average velocity over an interval is the plain average of the endpoints.

\[ v_{\text{avg}} = \frac{v_0 + v}{2} \]

The graphical answer is much more memorable than the algebraic one, so reach for the graph first.

With constant acceleration the velocity graph is a straight line. The displacement is the area under it, and the area under a straight line from v-nought to v is a trapezoid. A trapezoid's area is its width times the average of its two parallel sides — which is exactly the width times the midpoint of the two velocities. So the shortcut is not a physics fact, it is a fact about trapezoids.

Bend that line and the argument collapses immediately. If the velocity curves, the object spends unequal amounts of time near each speed, and the two endpoints stop being a fair summary of the interval.

The numerical example makes the size of the error clear. Take v equals t squared from zero to two seconds. The endpoint average gives two metres per second. The true average — displacement over time — is four thirds. The shortcut overstates it by fifty percent, because the particle spends most of the interval crawling along at low speed and only briefly reaches four.

So when you see a curved velocity graph and someone averages the endpoints, that is a real error and not a rounding issue.

30. Free Fall

Section

Section 4

31. Free fall: one acceleration, no exceptions

Concept

\[ a_y = -g, \qquad g = 9.8\,\text{m/s}^2 \]

Free fall means gravity is the only force acting. It does not mean falling — a ball thrown upward is in free fall the entire time it is in the air, including on the way up and at the apex.

The mass has cancelled out. Why that happens is a question for the next deck, when Newton's second law arrives; for now, take it as an experimental fact that costs a heavy object and a light one exactly the same time to fall.

On sign conventions: g is a positive number, 9.8. The minus sign belongs to the acceleration, and only if you chose up as positive. Writing g = -9.8 and then also writing a = -g is a double negative that has caused more lost marks than any other habit in this unit.

Two pieces of vocabulary and one sign convention, and this whole section becomes routine.

First, free fall does not mean falling. It means gravity is the only force acting. A ball thrown upward is in free fall from the instant it leaves your hand — on the way up, at the apex, and on the way down. All three phases have the same acceleration. This is why you never need to split a free-fall problem at the apex.

Second, the mass has cancelled out. The hammer and the feather in the vacuum chamber fall identically, and no property of the object appears in the acceleration at all. Why that happens is genuinely deep and is a question for the next deck, when Newton's second law arrives. For now take it as an experimental fact — and it is one you can watch on video from the Apollo 15 lunar surface.

Third, the sign convention, which is where marks actually get lost. The letter g stands for the positive number 9.8. The minus sign belongs to the acceleration and appears only if you chose up as positive. Writing g equals minus 9.8 and then also writing a equals minus g gives you a positive acceleration and a ball that flies away from the Earth. Pick g positive, always, and let the axis supply the sign.

32. A stone dropped from a bridge

Worked example

A stone is dropped from rest and hits the water 2.5 s later. How high is the bridge, and how fast is the stone moving on impact?

\[ \Delta y = v_0 t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(-9.8)(2.5)^2 = -30.6\,\text{m} \]

\[ v = v_0 + at = 0 + (-9.8)(2.5) = -24.5\,\text{m/s} \]

The negative answers are not errors to be tidied away. They are the coordinate system reporting back that the stone moved downward, which is what you told it to mean when you chose up as positive.

The translation from English to physics is the part to watch here. Dropped means the initial velocity is zero — that one word is the entire piece of given information, and problems rely on you catching it.

With up chosen as positive and the release point as the origin, the displacement comes out negative. That is not a mistake to be tidied away before writing the answer. It is the coordinate system correctly reporting that the stone finished below where it started. The bridge height is the magnitude of that number, and it is worth writing a sentence saying so rather than silently dropping the sign.

The impact velocity is also negative, for the same reason, and again the sign is information: it says downward. If the question asks for speed, report 24.5 metres per second. If it asks for velocity, keep the sign.

The verification is the useful habit here. The no-time equation gives the same impact speed by an entirely different route, using the displacement rather than the elapsed time. When two independent calculations agree, you are done worrying.

33. A ball thrown straight up

Worked example

A ball is thrown upward at 15 m/s from a height of 1.8 m. Find the maximum height above the ground and the total time in the air.

\[ 0 = (15)^2 + 2(-9.8)\Delta y \;\Rightarrow\; \Delta y = \frac{225}{19.6} = 11.5\,\text{m} \]

\[ y_{\text{max}} = 1.8 + 11.5 = 13.3\,\text{m} \]

\[ 0 = 1.8 + 15t - 4.9t^2 \]

\[ t = \frac{-15 - \sqrt{15^2 + 4(4.9)(1.8)}}{-9.8} \approx 3.18\,\text{s} \]

The upward and downward halves take equal times only when the launch and landing heights are equal. Here the ball has 1.8 m of extra falling to do, so the trip down is the longer one.

Two features of this problem are the ones examiners like, and both are easy to miss.

The first is that the apex condition is unstated. Nowhere does the problem say the vertical velocity is zero at the top — you have to know that and supply it. This is extremely common: maximum height problems always give you one fewer piece of data than you need, and the missing piece is always v equals zero at the peak.

The second is the asymmetry. Because the ball was launched from 1.8 metres up and lands at ground level, it has extra falling to do, and the trip down takes longer than the trip up. The neat symmetry you may have learned holds only when the launch and landing heights are the same. Here it does not, which is why the total time comes from solving a quadratic rather than from doubling the rise time.

And about that quadratic: it has two roots, and one of them is negative. That root corresponds to a time before the throw at which the ball would have been at ground level, had it already been in flight. It is mathematically real and physically meaningless, so discard it — but discard it deliberately, with a sentence explaining why, rather than silently.

34. At the highest point the acceleration is zero

Trap

35. Dropped versus fired horizontally

Prediction

This is the single most useful idea in two-dimensional kinematics, and it is worth stating as a slogan: horizontal velocity does not affect vertical fall time. The next section makes it precise.

This experiment is the foundation of everything in the two-dimensional section, and it is genuinely surprising the first time you see it done.

Both balls fall at the same rate. The one fired sideways travels much further and lands much further away, but it hits the floor at exactly the same instant as the one simply dropped. The horizontal motion has no effect whatsoever on the vertical timing.

The reason sits in the equations rather than in intuition. The vertical acceleration is minus g for both balls, and both start with zero vertical velocity. The vertical equations are therefore identical, and they contain no reference at all to the horizontal velocity. There is no term through which a sideways motion could influence the fall.

Say the slogan out loud a few times, because it will do a lot of work in the next section: horizontal velocity does not affect vertical fall time. Every projectile problem you meet is going to be solved by taking that seriously and treating the two directions as separate problems.

36. How fast are you moving after one second of falling?

Estimation

Estimation questions appear on the AP exam and are worth practising, because a wrong order of magnitude is usually a wrong physical picture.

Keep these two benchmarks in your head: after 1 s, about 10 m/s and 5 m. After 2 s, about 20 m/s and 20 m. Any answer that badly violates them deserves a second look before you write it down.

Note how differently the two grow. The speed doubles from 1 s to 2 s; the distance quadruples. That is the difference between a linear and a quadratic function, and it is visible in every free-fall problem you will meet.

Estimation questions appear on the AP exam, and they are not busywork. A wrong order of magnitude almost always means a wrong physical picture, so these are diagnostic.

The two benchmarks to memorise are: after one second, about ten metres per second and about five metres. After two seconds, about twenty metres per second and about twenty metres. Being able to produce those without a calculator will let you sanity-check any free-fall answer instantly.

The trap is that the speed and the distance are not the same number. Speed grows as g t, distance as one half g t squared, and that factor of one half is what people drop under time pressure. After one second the speed is nearly ten while the distance is only about five.

Look at how differently they grow, too. From one second to two, the speed doubles but the distance quadruples. That is the visible difference between a linear and a quadratic function, and once you have noticed it in free fall you will start seeing it everywhere — in stopping distances, in projectile ranges, and in the energy chapter.

37. Reading Motion off a Graph

Section

Section 5

38. The slope-and-area ladder

Concept

Two habits worth building. First, always check the vertical axis label before interpreting a graph — an identical-looking straight line means constant velocity on one graph and constant acceleration on another. Second, area below the axis is negative area.

This table is the whole of graphical kinematics, and it is worth being able to reconstruct it rather than memorise it.

Slope takes you down the ladder because slope is a rate of change, and each rung is the rate of change of the one above. Area takes you back up because area is accumulation, and integrating is accumulating. Both directions are the same two operations you already met algebraically.

The first habit worth building is to check the vertical axis label before interpreting any graph. A straight line sloping upward means constant velocity on a position graph and constant acceleration on a velocity graph — completely different physical situations that look identical on the page. Exam questions exploit this constantly.

The second is to remember that area below the axis is negative area. It is easy to see a shape and start computing its size without noticing which side of the axis it is on, especially when the graph crosses over.

One entry in the table is deliberately unhelpful: the area under a position graph. It has units of metre-seconds and no standard physical meaning. If a question seems to be asking for it, re-read — you have probably misidentified the axis.

39. Match each position graph to its velocity graph

Matching

The reliable method is to read the slope at three moments — the start, the middle, the end — and sketch those three values before worrying about the shape between them.

There is a reliable method for this kind of question that does not require you to recognise shapes, and it is worth using even when the answer looks obvious.

Read the slope of the position graph at three moments — the start, the middle, and the end. You now have three values for the velocity. Sketch those three points on a fresh set of axes and join them. That sketch is the velocity graph, and it will be right even for curves you have never seen before.

Applying it here: the straight line has one fixed slope throughout, so its velocity is a horizontal line. The upward-bending curve starts flat and gets steadily steeper, so its velocity starts at zero and rises. The downward-bending curve starts flat and then falls ever more steeply, so its velocity starts at zero and goes increasingly negative.

The common error is to match by visual resemblance — pairing a curve with a curve because they look alike. Slopes do not work that way. A parabola's slope is a straight line, so a curved position graph usually pairs with a straight velocity graph, and the resemblance is actively misleading.

40. Which graph feature tells you what

Discrimination

Every one of these features means something specific. Sort them by which graph you would be looking at when the feature matters.

Items c and d are the pair that catches people. Crossing the axis is a dramatic event on either graph — but on one it means back where you started and on the other it means turned around, and those are entirely different claims.

Every item in this sort is a true statement. The question is only ever which graph it is true of, and that is what makes it a good diagnostic.

The key is to keep asking what the vertical axis measures. On a position graph, the vertical axis is where the object is — so flat means not moving, crossing the axis means back at the origin, and a peak is a turning point. On a velocity graph, the vertical axis is how fast and which way — so flat means the velocity is not changing, crossing the axis means the direction reverses, the slope is the acceleration, and the area is the displacement.

Items c and d are the pair that catches almost everyone. Crossing the horizontal axis is a dramatic-looking event on either graph, but it means completely different things: on a position graph it means the object has returned to the origin, and on a velocity graph it means the object has turned around. An object can turn around a long way from the origin, and it can pass through the origin at full speed without turning at all.

If you find yourself unsure on a graph question, write the axis label at the top of your working before anything else. It resolves most of these ambiguities immediately.

41. Reading a piecewise velocity graph

Worked example

\[ \Delta x = 16 + 24 + 8 - 2 = 46\,\text{m}, \qquad \text{distance} = 16 + 24 + 8 + 2 = 50\,\text{m} \]

Piecewise graphs look intimidating and are actually the easiest kind, because every segment is a straight line and every area is a triangle or a rectangle. No calculus is needed at all.

Work segment by segment, and for each one take both readings: the slope, which gives you the acceleration during that segment, and the area, which gives you the displacement during it. Then combine.

The important moment is at nine seconds, where the line crosses the axis. That is where the object reverses direction, and it means the last segment contributes a positive area before the crossing and a negative area after it. Splitting there is not optional.

The check at the end is worth understanding rather than just performing. The distance exceeds the displacement by four metres, which is exactly twice the two metres of backtracking. That factor of two is the signature of a single reversal: the backtracked distance gets subtracted once in the displacement and added once in the distance, so the gap between them is double it. If your two answers differ by something that is not twice a backtracked amount, recheck the signs.

42. Graph reading

Check

A position-time graph is a parabola opening downward, with its vertex at t = 3 s.

The vertex of a position-time graph is a specific and much-tested feature, so it is worth being able to read it instantly.

At the vertex the tangent is horizontal, so the slope is zero, so the velocity is zero. And because the parabola opens downward, the slope is positive before the vertex and negative after it — the object was moving one way and is now moving the other. That is a turning point.

Option C is the tempting wrong answer and worth dismantling. A parabolic position graph corresponds to constant, non-zero acceleration everywhere: the curvature is what encodes it, and the curvature is the same at the vertex as anywhere else. The velocity vanishes there; the acceleration does not. This is the apex trap in graphical clothing.

Option A is a different confusion, between the graph's height and the graph's crossing point. Being at the origin means the curve touches the horizontal axis, which is unrelated to where its peak is — a parabola can peak far above the axis and never touch it at all.

43. Motion in Two Dimensions

Section

Section 6

44. The position vector and the unit vectors

Concept

\[ \vec{r}(t) = x(t)\,\hat{\imath} + y(t)\,\hat{\jmath} \]

The unit vectors i-hat and j-hat are constants — they have length one and never change direction. That is what makes the next slide's claim work: differentiating a vector means differentiating each component and leaving the unit vectors alone.

\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{\imath} + \frac{dy}{dt}\hat{\jmath} \qquad \vec{a} = \frac{d\vec{v}}{dt} \]

Everything from Section 1 now applies twice over. The ladder from position to velocity to acceleration runs independently in each component.

The notation on this slide is doing something quite specific, and once you see it the two-dimensional work becomes much less mysterious.

A position vector is built from two ordinary numbers — the coordinates — and two unit vectors that carry the directional information. The crucial property of i-hat and j-hat is that they are constants: they always have length one and they never change direction, no matter what the object does.

That constancy is exactly what makes differentiation easy. When you differentiate the position vector, the unit vectors come out of the derivative untouched, and you are left differentiating the two coordinate functions separately. So finding a velocity vector is nothing more than finding two ordinary derivatives and reattaching the hats.

This is why everything from the first section of this deck now applies twice over rather than needing to be rebuilt. The ladder from position to velocity to acceleration runs independently down each component, using exactly the calculus you already have.

45. x and y do not talk to each other

Concept

This is the load-bearing idea of two-dimensional kinematics, and it follows directly from the component form above.

\[ a_x = \frac{d^2x}{dt^2} \qquad a_y = \frac{d^2y}{dt^2} \]

There is no term in the x equation that mentions y, and none in the y equation that mentions x. The two are completely separate one-dimensional problems that happen to be running on the same clock.

The only thing the two columns share is t. That is the bridge: solve one column for the time, then hand that time to the other. Nearly every projectile problem is built on exactly that handoff.

This is the load-bearing idea of two-dimensional kinematics, and the argument for it is embarrassingly short: look at the two equations and notice that neither one mentions the other's variable.

There is no term in the x-equation containing y, and none in the y-equation containing x. So the horizontal motion cannot influence the vertical motion and vice versa. They are two entirely separate one-dimensional problems that happen to be running on the same clock.

For a projectile, that separation is dramatic. The horizontal column has zero acceleration, so it is the simplest possible motion — constant velocity, distance equals speed times time. The vertical column has minus g and gets the full apparatus of the big four. Two very different problems, solved with different tools, in the same question.

The bridge between the columns is t, and only t. The standard structure of every projectile problem is therefore: solve one column for the time, then hand that time to the other column. Recognising which column to solve first is nearly the whole skill — usually it is whichever one has the more complete information.

46. Two shadows, two separate movies

Picture it

Imagine a projectile lit by two lamps: one directly overhead casting a shadow on the ground, one off to the side casting a shadow on a wall.

The ground shadow is the x-motion: evenly spaced dots, because there is no horizontal acceleration. The wall shadow is the y-motion: dots bunching together near the top, because the vertical velocity is shrinking to zero there.

Neither shadow shows a parabola. The parabola is only what you see when you watch both at once — it is a picture of the two motions combined, not a thing the projectile is separately doing.

This picture is the one to hold onto when the algebra of projectiles starts feeling abstract.

Imagine two lamps: one directly overhead casting the projectile's shadow onto the ground, one off to the side casting it onto a wall. Each shadow shows you one component of the motion, stripped of the other.

The ground shadow moves at a perfectly constant speed. Its dots are evenly spaced, because there is no horizontal acceleration and never was. If you only watched the ground shadow, you would see something as boring as a marble rolling on a table.

The wall shadow is a different film entirely. It rises, slows, stops, reverses, and comes back down, with the dots bunching up near the top where the vertical velocity is smallest. If you only watched that shadow, you would see a ball thrown straight up.

The crucial observation is that neither shadow is a parabola. The parabola only exists when you watch both at once. It is not something the projectile is separately doing — it is what two simple motions look like when superimposed. Once that lands, the fact that the two columns can be solved independently stops feeling like a trick.

47. Projectile launched from the ground

Worked example

A ball is launched from ground level at speed v0 and angle theta above the horizontal. Derive its trajectory, then find where it lands.

\[ v_{0x} = v_0\cos\theta \qquad v_{0y} = v_0\sin\theta \]

\[ x(t) = v_0\cos\theta \cdot t \qquad y(t) = v_0\sin\theta \cdot t - \tfrac{1}{2}gt^2 \]

\[ t\left(v_0\sin\theta - \tfrac{1}{2}gt\right) = 0 \;\Rightarrow\; t = 0 \;\text{or}\; t = \frac{2v_0\sin\theta}{g} \]

\[ R = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g} = \frac{v_0^2\sin 2\theta}{g} \]

This derivation is worth doing once by hand even though the results are on every formula sheet, because the structure is the structure of every projectile problem you will ever solve.

Notice that the trigonometry happens exactly once, at the very start, when the launch velocity is resolved into components. After that the angle never appears again until the final answer. If you find yourself reaching for sine and cosine in the middle of a projectile problem, something has gone wrong with the setup.

The time of flight comes from asking when the vertical position returns to zero, and it is worth factoring rather than using the quadratic formula. Factoring makes the two roots meaningful: t equals zero is the launch itself, and the other root is the landing. Blindly applying the quadratic formula gets the same numbers but hides that structure.

Then comes the handoff — the moment that defines these problems. The time was found in the vertical column and is now used in the horizontal one, because t is the only thing the two columns share.

And check the result at the end. At zero degrees the range is zero, which is obviously right. At forty-five degrees sine of two theta hits its maximum of one, giving the longest throw — which matches what anyone who has thrown a ball already believes. A formula that survives both checks is probably right.

48. Range, time of flight, and maximum height

Concept

\[ t_{\text{flight}} = \frac{2v_0\sin\theta}{g} \qquad R = \frac{v_0^2\sin 2\theta}{g} \qquad H = \frac{v_0^2\sin^2\theta}{2g} \]

These three hold only for launch and landing at the same height. That restriction is the most common reason a memorised range formula produces a wrong answer — a ball thrown off a cliff obeys none of them.

Because sin(2 theta) equals sin(180 degrees minus 2 theta), complementary angles give identical ranges. A 30-degree throw and a 60-degree throw land in the same place — the steep one just takes longer and goes higher to get there.

These three formulas are genuinely useful and genuinely dangerous, and the danger is entirely in the restriction printed above them.

They assume the projectile lands at the same height it was launched from. That assumption is buried in the derivation — it is where we set y back to zero — and it is invisible in the final formulas. A ball thrown off a cliff, or into a basket, or from a table, obeys none of these three.

So the practical rule is: if the launch and landing heights differ at all, throw these formulas away and go back to the two columns. It is not much extra work and it is always correct.

Within their domain, though, the complementary-angle result is genuinely elegant. Because sine of two theta equals sine of one-eighty minus two theta, a thirty-degree throw and a sixty-degree throw land in exactly the same place. They get there very differently — the steep one goes much higher and takes much longer — but the landing point is identical. That symmetry is visible in the figure and is a favourite of multiple-choice writers.

49. Turn the launch angle

Tweak it

Hold the launch speed fixed and sweep the angle. Watch which quantity peaks where.

The asymmetry between the three quantities is the point. Range is a compromise between how long you stay up and how fast you go sideways, and 45 degrees is where the product of those is largest. Height and flight time have no such competition, so they just keep rising with the angle.

The point of sweeping the angle is to notice that the three quantities behave completely differently, which is not obvious from the formulas alone.

The range peaks at forty-five degrees and falls off symmetrically either side. The maximum height and the flight time do not peak at forty-five at all — they keep climbing all the way to ninety degrees, where the projectile goes straight up.

The reason is a competition that only the range has to worry about. Range is the product of horizontal speed and time aloft. Raising the angle buys you more time but costs you horizontal speed, and forty-five degrees is where that product is largest. Height and flight time face no such trade-off; they depend only on the vertical component, so more angle is simply more of both.

This is a good habit to generalise: when a formula has a maximum somewhere in the middle of its range, look for the two competing effects. There are almost always exactly two, and identifying them tells you more than the calculus that located the maximum.

50. At the top of its arc, the projectile's velocity is zero

Trap

51. Launched horizontally off a cliff

Worked example

A ball rolls off a 45 m cliff at 12 m/s horizontally. Find the time to land, the horizontal distance travelled, and the impact speed.

\[ -45 = -\tfrac{1}{2}(9.8)t^2 \;\Rightarrow\; t = \sqrt{\frac{90}{9.8}} = 3.03\,\text{s} \]

\[ x = (12)(3.03) = 36.4\,\text{m} \]

\[ v_y = -(9.8)(3.03) = -29.7\,\text{m/s} \qquad v = \sqrt{12^2 + 29.7^2} = 32.0\,\text{m/s} \]

Notice that the 12 m/s never once entered the time calculation. That is the horizontal-vertical independence from three slides ago, doing real work.

This is the cleanest possible demonstration of the two-column method, because one of the columns has been stripped to almost nothing.

Launched horizontally means the initial vertical velocity is zero. So the vertical column is just an ordinary drop — exactly the stone-off-the-bridge problem from earlier in this deck, unchanged. The fall takes as long as it would if the ball had simply been released from the cliff edge.

Watch what does the work in the time calculation: only the height and g. The twelve metres per second never appears. That is horizontal-vertical independence doing something concrete rather than being asserted.

The impact velocity is then a vector sum, and the two components must be combined with Pythagoras rather than added. The horizontal component is still twelve, untouched all the way down; the vertical component has grown to nearly thirty.

The independent check at the end is worth copying into your own work. Using the no-time equation to get the vertical speed uses the height rather than the time, so it is a genuinely separate calculation. When it agrees, both the time and the speed are confirmed at once.

52. Push the launch angle to its limits

Edge cases

Formulas are best understood at their edges. Take the range formula and push the angle to each extreme.

This is a habit worth carrying into every formula you meet this year: set each variable to zero and to infinity, and check that the formula says something you already believe. It catches algebra errors faster than re-checking the algebra.

Testing a formula at its extremes is one of the highest-value habits available to you this year, and this is a good place to practise it.

Push the angle toward zero. The horizontal speed is almost the full launch speed, which sounds promising — but the vertical speed is nearly nothing, so the projectile is back on the ground almost immediately. Plenty of speed, no time to spend it.

Push toward ninety and the situation inverts. The flight time is as long as it can be, but there is essentially no horizontal velocity to convert that time into distance. Plenty of time, no speed to spend it.

Both ends give a range of zero, and they get there for opposite reasons. Forty-five degrees is where the two effects balance. Understanding the range formula this way is much more durable than memorising it, and it means you could reconstruct the answer if the formula slipped your mind.

The habit generalises: take any formula you are unsure of, send each variable to zero and to infinity, and check that what comes out is something you already believe. It catches algebra errors faster than re-checking the algebra does.

53. Why long jumpers do not take off at 45 degrees

Real world

The range formula says 45 degrees is optimal. Film of elite long jumpers shows take-off angles closer to 20 degrees. Both are correct.

This is a good example of a formula being completely correct and completely useless at the same time, and the reason is an assumption that is easy to miss.

The range formula treats the launch speed as fixed while the angle varies. For a cannon that is a fair description — the powder does not care which way the barrel points. For a human being it is badly false.

To leave the ground at forty-five degrees, a jumper would have to convert a large fraction of a very fast run-up into vertical velocity during the fraction of a second the foot is on the ground. Legs cannot deliver that impulse. Attempting a steeper take-off therefore costs an enormous amount of speed, and the loss more than cancels the gain from the better angle.

The real optimisation is over a launch speed that falls as the angle rises, and its maximum sits far lower — experimentally around twenty to twenty-two degrees, which is what film of elite jumpers actually shows.

The transferable lesson: a formula's optimum is only optimal within the formula's assumptions. Before trusting one about the real world, ask what it is quietly holding constant, and whether the real system can actually hold that thing constant.

54. Relative Motion

Section

Section 7

55. Velocities add — if you chain the subscripts

Concept

\[ \vec{v}_{A\,\text{rel}\,C} = \vec{v}_{A\,\text{rel}\,B} + \vec{v}_{B\,\text{rel}\,C} \]

Read the subscripts as a chain: A relative to B, then B relative to C, gives A relative to C. The inner labels must match and then cancel, exactly like units cancelling in a conversion.

\[ \vec{v}_{B\,\text{rel}\,A} = -\,\vec{v}_{A\,\text{rel}\,B} \]

And reversing the subscripts flips the sign. If the train moves at 30 m/s relative to the platform, the platform moves at 30 m/s backwards relative to the train — which is exactly what it looks like out of the window.

Relative velocity problems are almost entirely a bookkeeping exercise, and the subscript chain is the bookkeeping system that makes them safe.

Read the subscripts as a chain: A relative to B, then B relative to C, gives A relative to C. The inner labels have to match, and when they do they cancel — exactly like units cancelling in a unit conversion. If your inner labels do not match, you have set the problem up wrongly and no amount of careful arithmetic will rescue it.

The reversal rule is the other half. Swapping the subscripts flips the sign of the vector. If the train moves at thirty metres per second relative to the platform, then the platform moves at thirty metres per second backwards relative to the train — which is exactly what it looks like out of the window, and is a good reminder that neither point of view is more correct than the other.

The walkway picture is worth keeping because it makes the addition physical. You are walking at your own speed relative to the moving floor, and the floor is moving relative to the building. Your speed relative to the building is the two arrows laid head to tail, which is all vector addition ever is.

56. Rowing across a moving river

Worked example

A river flows east at 3.0 m/s. A boat can row at 4.0 m/s relative to the water, and the river is 80 m wide. The rower points the boat due north. Where does it land, and how long does the crossing take?

\[ \vec{v}_{BG} = 4.0\,\hat{\jmath} + 3.0\,\hat{\imath} \]

\[ t = \frac{80}{4.0} = 20\,\text{s} \]

\[ \Delta x = (3.0)(20) = 60\,\text{m downstream} \]

The counter-intuitive part is the time: the current does not slow the crossing at all. To land directly opposite, the rower would have to aim upstream, and that would cost time — the northward component would drop to 2.65 m/s and the crossing would take 30 s.

The counter-intuitive result in this problem is the crossing time, so make sure you see why it comes out the way it does.

The current is entirely eastward. The gap to be closed is entirely northward. An eastward velocity contributes precisely nothing to closing a northward gap, so the crossing time depends only on the rowing speed and the river width. Twenty seconds, current or no current.

This is the same independence you met with projectiles, wearing different clothes. Perpendicular components do not influence one another, whether the perpendicular pair is horizontal-and-vertical or across-and-downstream.

The current does have an effect, of course — it sweeps the boat sixty metres downstream. It changes where you land, not when you land.

And the follow-up is worth thinking through, because it is a standard exam extension. To land directly opposite, the rower must aim upstream, and then part of the four metres per second is being spent fighting the current rather than crossing. The northward component drops to about two point six five, and the crossing now takes thirty seconds instead of twenty. Landing where you want costs you time; that trade-off is the physics of the problem.

57. Relative velocity

Check

Two cars approach each other on a straight road. Car A moves east at 25 m/s and car B moves west at 15 m/s, both measured relative to the road.

Approaching vehicles are the case where intuition and arithmetic most often disagree, so it is worth doing this one slowly with the subscript chain rather than by feel.

Set east as positive. Car B relative to the road is minus fifteen. The road relative to car A is minus twenty-five — note the reversal, since A relative to the road was plus twenty-five. Chaining them gives minus forty, so car B approaches at forty metres per second westward as seen from A.

The physical statement is that vehicles moving toward each other close on one another at the sum of their speeds. That is why head-on collisions are so much more severe than rear-end collisions at the same speeds: the closing speed is the sum rather than the difference.

The tempting wrong answer subtracts the two speeds, which is the right move for two cars travelling in the same direction and the wrong move here. The other tempting error gets the size right and the direction backwards, which is what happens when the subscripts get reversed somewhere in the chain. Both are avoided by writing the chain down rather than reasoning verbally.

58. When the Acceleration Will Not Hold Still

Section

Section 8

59. Three flavours of variable acceleration

Concept

This section is where Physics C leaves Physics 1 behind. The big four are gone; calculus is all you have, and which technique you use depends on what the acceleration is a function of.

\[ a = \frac{dv}{dt} = v\frac{dv}{dx} \]

Both forms of that identity are on your equation sheet in spirit, and knowing when to reach for the second one is what the harder free-response questions are testing.

The diagnostic question is always the same: what variable is on the right-hand side? Match the technique to that, and these problems become mechanical.

This is where Physics C genuinely departs from Physics 1, and the good news is that the whole section runs on a single diagnostic question.

The question is: what variable is the acceleration a function of? Not what the problem is about, not what is being asked for — just what appears on the right-hand side. That answer picks the technique, and once the technique is picked the rest is routine calculus.

If a depends on time, you can integrate directly, because dv/dt equals a of t separates immediately. If a depends on velocity, you cannot integrate over time — you do not yet know v of t, which is the very thing you are solving for — so you separate variables instead, putting all the v on one side and all the t on the other. If a depends on position, you use the identity a equals v dv/dx to trade away the time variable entirely.

That identity is worth memorising in both directions. It is on the equation sheet in spirit and it is the key to a large fraction of the harder free-response questions, in this unit and again in energy and oscillations.

60. Air resistance: a = -kv

Worked example

A boat of mass m moving at v0 shuts off its engine. The water resists with a force proportional to speed, giving a = -kv. Find v(t).

\[ \frac{dv}{dt} = -kv \quad \Rightarrow \quad \frac{dv}{v} = -k\,dt \]

\[ \int_{v_0}^{v}\frac{dv'}{v'} = \int_0^t -k\,dt' \quad \Rightarrow \quad \ln\frac{v}{v_0} = -kt \]

\[ v(t) = v_0 e^{-kt} \]

Read the physics off the result: the boat never actually stops. Integrating v(t) from zero to infinity gives a finite total distance of v0/k, so it coasts a definite distance while taking forever to do it.

This is the standard separation-of-variables problem, and it appears in some form on a great many Physics C exams, so the pattern is worth owning.

Recognise the flavour first: acceleration depends on velocity. That immediately rules out direct integration over time, because integrating minus k v with respect to t requires knowing v as a function of t, which is what you are trying to find. Circular. So separate instead.

Once separated, the integral of dv over v is a natural logarithm, which is why exponentials turn up in every drag problem you will ever meet. Attaching the limits — v-nought to v on one side, zero to t on the other — is cleaner than carrying a constant of integration, and it removes a step where errors creep in.

The result is exponential decay, and the physics in it is worth reading off. The boat never actually stops. Its speed halves, halves again, and keeps halving forever, approaching but never reaching zero.

Yet it does not travel infinitely far. Integrating the velocity from zero to infinity gives a finite distance, v-nought over k. So the boat coasts a definite, calculable distance while taking literally forever to do it. That combination surprises people, and it is a good illustration of why an infinite time does not imply an infinite distance.

61. The v dv/dx trick

Concept

When acceleration is a function of position, time is a nuisance variable — it appears nowhere in the given information and is not usually wanted in the answer.

\[ a(x) = v\frac{dv}{dx} \quad \Longrightarrow \quad \int_{x_0}^{x} a(x')\,dx' = \int_{v_0}^{v} v'\,dv' = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \]

The left side is an integral over position, which is what you were given. The right side is a difference of v squared terms. Time has been eliminated entirely.

You have met this shape before. Multiply through by the mass and the left side is the work done, while the right side is the change in kinetic energy. The work-energy theorem is this identity with units attached — which is why energy methods and v dv/dx always give the same answer.

Practical rule: if a problem gives you acceleration or force as a function of position and asks for a speed, this is the fastest route, and it works whether or not the force is conservative.

This identity solves a specific irritation: you have been given acceleration as a function of position, and time appears nowhere in the problem and is not wanted in the answer.

Writing a as v dv/dx eliminates time from the mathematics entirely. Separate it and integrate, and the left side is an integral over position — which is what you were given — while the right side is a difference of v-squared terms, which is what you want.

Now look hard at the shape of that result: one half v squared minus one half v-nought squared on one side, an integral of a with respect to x on the other. Multiply both sides by the mass. The right side becomes the integral of force over distance, which is the work done. The left becomes the change in kinetic energy.

So you have just derived the work-energy theorem without intending to. This is why energy methods and the v dv/dx technique always agree — they are the same mathematics with different labels, and you may use whichever is more convenient.

The practical rule: given force or acceleration as a function of position and asked for a speed, this is the fastest route. It works whether or not the force is conservative, which is more than can be said for potential-energy methods.

62. Acceleration given as a function of position

Worked example

A particle starts at rest at x = 0 and experiences a = 6x squared, with x in metres. Find its speed at x = 2 m.

\[ v\frac{dv}{dx} = 6x^2 \]

\[ \int_0^v v'\,dv' = \int_0^2 6x'^2\,dx' \]

\[ \tfrac{1}{2}v^2 = 2x^3\Big|_0^2 = 16 \]

\[ v = \sqrt{32} = 5.66\,\text{m/s} \]

Note what was never needed: how long any of this took. That question would require a much harder integration, and the problem was careful not to ask it.

This problem is short, and its value is entirely in the recognition step rather than the algebra.

The acceleration is given as six x squared — a function of position. Integrating that over time is impossible without knowing x of t, which is not available and would be hard to get. So use the v dv/dx identity, which converts the problem into an integral over position that you can actually evaluate.

Attach limits rather than carrying constants: velocity runs from zero to v while position runs from zero to two. Both integrals are then elementary power-rule work, and the answer falls out in one line.

The consistency check at the end is the habit worth copying. The acceleration is positive everywhere past the origin, so the particle only ever speeds up, which means the speed at two metres must exceed the speed at one metre. Running the same integral to x equals one gives two metres per second, comfortably less than five point six six. The answer behaves the way the physics says it must.

Notice too what the problem carefully did not ask: how long any of this took. That question would require a much nastier integration, and its absence is a hint about which technique was intended.

63. Terminal velocity: acceleration that erases itself

Anomaly

Drop an object through real air and the acceleration is not constant. The drag force grows with speed, so the net force shrinks as the object speeds up.

\[ a = g - \frac{b}{m}v \]

Set the acceleration to zero and read off the answer: v_terminal = mg/b. Notice the mass is back — heavier objects have higher terminal velocities, which is why the feather and the hammer part company as soon as you let the air back in.

Real falling objects do not have constant acceleration, and this is the standard model of why.

The drag force grows with speed. So as the object speeds up, drag eats further into gravity, and the net acceleration shrinks. The faster it goes, the less it accelerates — a self-limiting process.

Eventually the drag term has grown all the way to g, the acceleration reaches zero, and the velocity stops changing. That equilibrium speed is the terminal velocity, and you find it by the simplest possible method: set the acceleration to zero and solve. No integration required.

The approach to it is asymptotic — the object gets arbitrarily close and never quite arrives, which is the same exponential behaviour as the coasting boat a few slides ago. Both come from a rate proportional to how far you are from equilibrium.

And notice what has come back into the physics: the mass. Terminal velocity is mg over b, so heavier objects fall faster in air. That is why the feather and the hammer only behave identically in the vacuum chamber — restore the air and the feather's tiny mass gives it a tiny terminal velocity almost immediately.

64. Choosing the technique by what a depends on

Pattern

Every kinematics problem in this course can be routed by two questions.

1. Is the acceleration constant? If yes, use the big four and stop reading.
2. If not, what is a a function of? That answer picks the technique.

Notice how few of the rows are the big four. In Physics 1 that row was the entire subject; here it is one special case among several, and the exam knows it.

Two questions route every kinematics problem in this course, and this table is those two questions written out in full.

First: is the acceleration constant? If it is, use the big four, pick the equation missing the variable nobody mentioned, and you are done. If it is not, that entire toolkit is off the table and you must not reach for it out of habit.

Second, if it is not constant: what is the acceleration a function of? Time, velocity, or position — each gets a different technique, and the middle column of the table names it.

Look at the proportions in the table for a moment. In Physics 1 the first row was the entire subject. Here it is one case among many, and the exam knows exactly which rows separate the two courses. The variable-acceleration rows are where the free-response points live.

Spend your practice time proportionally. If you can already do constant-acceleration problems reliably, more of them will not raise your score; a dozen separation-of-variables problems will.

65. Three questions before you close the deck

Exit ticket

If any of the three gave you pause, the slide that fixes it is nearby: the sign trap in Section 1, the constant-of-integration trap in Section 2, and the apex trap in Section 6.

Retrieval beats re-reading by a wide margin, so answer these from memory before looking at the responses — including the ones you are sure of.

The first question is the sign trap. Opposite signs means the acceleration is fighting the motion, so the object is slowing down. It will slow to a stop and then start moving in the positive direction, at which point the signs agree and it begins speeding up.

The second is the constants question. Two constants, one after each integration, and each evaluated before the next integration begins. The order matters, not just the count.

The third is the apex trap in its two-dimensional form. The acceleration at the apex is 9.8 metres per second squared downward, exactly as everywhere else in the flight. Gravity is the only thing acting and it never changes during the flight.

If any of the three made you hesitate, go back to the slide that fixes it — the sign trap in Section 1, the constant-of-integration trap in Section 2, the apex trap in Section 6 — rather than re-reading the whole deck.

66. What you can do now

Recap

• Move between x, v and a by differentiating down the ladder and integrating back up, paying an initial condition each time
• Derive the constant-acceleration equations rather than memorising them, and say exactly when they are legal
• Choose the right one of the big four by spotting the variable nobody mentioned
• Read slopes and signed areas off position, velocity and acceleration graphs
• Split any two-dimensional problem into two independent columns joined only by t
• Handle a(t), a(v) and a(x) with the right integration technique for each

Next deck: forces. Everything here described motion without asking what caused it. Newton's laws supply the cause, and the acceleration you have been handed all deck becomes something you can predict from a free-body diagram.

Take stock of what has actually changed. You started this deck able to describe motion; you finish it able to compute motion from any description of how it is changing.

The single organising idea is the ladder. Position, velocity, acceleration, connected downward by differentiation and upward by integration, with an initial condition owed at every upward step. The four constant-acceleration equations are one special case of that ladder, not a separate subject.

The habits are worth as much as the physics. Draw the axis first. Solve symbolically before substituting. Check the extremes of every formula. Verify by an independent route when you can. None of these are about kinematics specifically — they will carry you through every remaining unit.

What comes next is the other half of mechanics. Everything here described motion without ever asking what caused it, which is why the acceleration always had to be handed to you in the problem statement. Newton's laws supply the cause, and from the next deck onward the acceleration is something you derive from a free-body diagram rather than something you are given.