Unit 1 of the algebra-based course, organised around a five-step attack on any motion problem rather than around theory. It fixes the five quantities and the sign rule that decides speeding up from slowing down, gives the one-step method for choosing between the four constant-acceleration equations by finding the quantity a problem never mentions, covers free fall including the acceleration at the top of a throw, resolves vectors with trigonometry, then treats projectiles as two independent problems joined only by a shared clock, and closes with relative velocity and a homework-facing summary. No calculus is used anywhere.
Subject: AP Physics 1 · 62 slides · diagram-first lesson
Open the interactive version of this deck · Homework for this lesson
Title
Unit 1
A repeatable attack on motion problems, from one dimension to projectiles
Objectives
Your school is still on the maths refresher, so this is the run-up rather than catch-up. The aim is not to read about motion — it is to leave with a method you can put on any homework problem, and the confidence that comes from the method working every time.
College Board, AP Physics 1: Algebra-Based Course and Exam Description Unit 1, Kinematics
Section
How to attack any problem
Concept
There are only four equations in this entire unit, and you will have them memorised within a week. That is not what separates students who find the homework easy from students who do not.
What separates them is having something to do when they read a problem and do not immediately see the answer. Without a method, a hard problem produces staring. With one, it produces a first line of work, and the first line usually reveals the second.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 2, problem-solving strategies
Picture it
Write these on the inside cover of your notebook. Use them even on the problems that look easy, because that is how they become automatic before the problems get hard.
Figure (svg): A five-step vertical flow from drawing the situation through listing knowns, naming the unknown, choosing the equation and checking the answer
Steps one to three take longer than the algebra and are where nearly all the marks are won. Skipping straight to an equation is the single most common cause of a wrong answer in this unit.
Prediction
Before we start, a guess about your own likely failure mode.
Predict first
On a typical kinematics homework, where do most lost marks come from?
Correct: Sign errors and choosing the wrong equation
Why: The equations are few and quickly learned, and calculators handle the arithmetic. What goes wrong is upstream: a direction chosen inconsistently, or three minutes spent on an equation that could never have worked because it needs a quantity the problem does not give. Both are fixed by steps one and four of the method, which is why they get a section each.
Section
One decision, made early
Concept
Every constant-acceleration problem is about five quantities. Getting the vocabulary exact matters because the exam wording is precise and the difference between two of these words is often the whole question.
| symbol | name | the distinction that catches people |
|---|---|---|
| v0 | initial velocity | at rest means this is zero, not that acceleration is zero |
| v | final velocity | final means at the end of the interval you chose, not at the end of the motion |
| a | acceleration | a rate of change of velocity, so it is nonzero whenever speed or direction changes |
| t | time interval | an elapsed time, not a clock reading |
| dx | displacement | a change of position, with a sign — not the distance travelled |
Distance and displacement come apart the moment something reverses. A ball thrown up and caught again has travelled a distance but has a displacement of zero, and problems exploit that on purpose.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 2.1
Picture it
This is the decision that prevents most sign errors, and it costs five seconds.
Figure (svg): A horizontal axis with a chosen positive direction, arrows showing positive and negative velocity, and the speeding-up rule stated beneath
Once you have drawn that arrow, every quantity in the problem gets its sign from it, including acceleration. Down is not automatically negative — it is negative only if you chose up as positive, and you should say so on the page.
Definition probe
The pair that problems exploit most often. A runner does a full lap of a 400 metre track.
Sort into buckets
Which quantity is being described?
Matching
Exam wording hides your third known inside an ordinary-sounding phrase. Learn the dictionary and the problems get shorter.
Match the pairs
Why: Every one of these supplies a number the problem never states as a number. Missing one is the usual reason a problem looks unsolvable - you have two knowns and need three, and the third is sitting in a phrase you read past.
Pattern
A short table you can apply mechanically until it becomes instinct.
| the situation | with up chosen positive | why |
|---|---|---|
| a ball rising | v positive, a negative | moving up, pulled down |
| a ball at the top | v zero, a negative | momentarily stopped, still pulled down |
| a ball falling | v negative, a negative | moving down, pulled down |
| a car speeding up forwards | v positive, a positive | same signs means speeding up |
| a car braking while moving forwards | v positive, a negative | opposite signs means slowing down |
Read the last two rows together. Whether an object speeds up or slows down is not about the sign of the acceleration on its own — it is about whether the two signs agree.
HyperPhysics, Mechanics — motion in one and two dimensions one-dimensional motion
Check
Take up as positive throughout.
Check your understanding
A ball is thrown upward and is currently moving up but slowing. What are the signs of its velocity and acceleration?
Answer: B
Why: It is moving up, so the velocity is positive under this choice of axis. It is slowing, so the acceleration must have the opposite sign to the velocity, making it negative. That matches gravity pulling downward the entire time.
Discrimination
Use the sign rule rather than intuition — two of these are deliberately counter-intuitive.
Sort into buckets
Is the object speeding up or slowing down?
Section
And how to choose in one step
Concept
Here they are. They apply only when the acceleration is constant, which is true in nearly every problem this unit sets, and you should get into the habit of confirming it before using them.
\[ v = v_0 + at \]
\[ \Delta x = v_0 t + \tfrac{1}{2}at^2 \]
\[ v^2 = v_0^2 + 2a\,\Delta x \]
\[ \Delta x = \tfrac{1}{2}(v_0 + v)\,t \]
College Board, AP Physics 1: Algebra-Based Course and Exam Description Unit 1 equations
Picture it
This is the observation that turns equation choice from a search into a single step, and it is worth more than anything else in this deck.
Figure (svg): The four kinematic equations listed with the quantity each one omits marked beside it
So the recipe is: list the five quantities, mark the one the problem neither gives you nor asks for, and use the equation that does not contain it. One pass, no trial and error.
Socratic
Worth understanding rather than just using, because it tells you when it will not work.
Discussion prompt
Why is there exactly one equation missing each quantity, and what does that tell you about how many things a problem must give you?
Hint: How many knowns does one equation need before it can be solved?
Answer:
Each equation relates four of the five quantities, so it can solve for one unknown if you know the other three.
That means a well-posed problem always gives you three of the five. If you can only find two, something is unstated — often the initial velocity being zero, or the object ending at rest.
So when the trick seems not to work, do not go looking for a fifth equation. Go looking for the third piece of given information hidden in a phrase like from rest, or dropped, or comes to a stop.
Worked example
A car starts from rest and accelerates uniformly at 3.0 metres per second squared for 8.0 seconds. How far does it travel?
Figure (svg): A velocity-time graph rising in a straight line from zero to twenty-four metres per second over eight seconds, with the triangular area beneath shaded
Draw and choose an axis
Why: A straight road, positive in the direction of travel. Nothing here is negative.
List the knowns with signs
Why: From rest gives an initial velocity of zero; that is the third known that makes the problem solvable.
\[ v_0 = 0, \quad a = 3.0, \quad t = 8.0 \]
Name the unknown
Why: The displacement.
Choose by what is missing
Why: The problem never mentions final velocity, so use the equation without it.
\[ \Delta x = v_0 t + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(3.0)(8.0)^2 \]
\[ \Delta x = 96 \text{ m} \]
Verify: with the graph area
Why: The velocity-time graph is a straight line from zero to 24 metres per second over 8 seconds, and the area under it is half of 8 times 24, which is 96 metres. Two independent routes agreeing is what makes an answer safe to write down.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 2.5
Fill the middle
A car moving at 25 metres per second brakes to a stop in 50 metres. Find the acceleration.
Fill in the blanks
The quantity never mentioned is time, so I use the equation v squared equals v0 squared plus 2 a dx, and the acceleration comes out negative 6.25 metres per second squared.
Why: Zero equals 625 plus 2a times 50, so 100a equals negative 625 and a is negative 6.25. The sign is negative because the car is moving in the positive direction and slowing, and that negative is part of the answer rather than an accident of the algebra.
Estimation
Number sense is what catches an answer that is wrong by a factor of ten, and it takes two seconds.
Predict first
A car doing about 30 metres per second brakes hard, at roughly 7 metres per second squared. Roughly how far to stop?
Correct: About 65 m
Why: The equation without time gives distance as the speed squared over twice the deceleration: 900 over 14, which is about 64 metres. That is around fifteen car lengths, which matches the stopping distances on a road sign. Any answer near 15 metres should feel wrong immediately at motorway speed.
Socratic
A genuinely useful consequence of one of the four equations, and the reason speed limits matter more than people think.
Discussion prompt
If the car above were doing 60 metres per second instead of 30, how much further would it take to stop, and why?
Hint: Which power of v appears in the equation that omits time?
Answer:
Four times as far, not twice. The stopping distance depends on the square of the speed, since the equation without time has v squared in it.
So 3600 over 14 is about 257 metres, against 64 metres at half the speed.
This is why the relationship between speed and stopping distance is not intuitive, and it is a standard exam question dressed up as a road-safety scenario.
Trap
A ball is dropped and hits the ground 3.0 seconds later. How fast was it going on impact?
Grab the equation with a squared velocity in it
Why: It has velocity in it, so it must be the one. But it needs a displacement, which the problem did not give.
Get stuck, then invent a height
Why: Time is wasted, and any number invented here makes the rest of the work wrong even if the algebra is perfect.
A ball is dropped and hits the ground 3.0 seconds later. How fast was it going on impact?
List first: dropped means initial velocity zero, a is 9.8 downward, t is 3.0
Why: Three knowns, so the problem is solvable.
Mark what is never mentioned
Why: Displacement is neither given nor asked for, so use the equation without it.
Solve in one line
Why: v equals 0 plus 9.8 times 3.0, which is 29.4 metres per second downward. The height was never needed, and going looking for it was the error.
Section
The one acceleration you always know
Concept
Free fall means the only force acting is gravity, so air resistance is being ignored. Every problem in this unit says or implies it, and it is worth noticing that it is an approximation rather than a fact.
Under that assumption the acceleration is the same for every object regardless of mass, directed downward, with a magnitude of about 9.8 metres per second squared. A feather and a hammer really do fall together in a vacuum, and mass never enters a kinematics equation.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 2.7
Picture it
The single most examined situation in this unit. Look at what happens at the top before reading on.
Figure (svg): A ball thrown upward shown at five moments with its velocity at each, and a panel noting that acceleration remains 9.8 downward even at the top
At the top the velocity is zero for an instant and the acceleration is completely unchanged. That is the question examiners ask most often, and the intuitive answer is wrong.
Anomaly
Sit with this one for a moment before answering.
Predict first
At the highest point of its flight, what is the ball's acceleration?
Correct: 9.8 metres per second squared downward
Why: Acceleration measures how fast velocity is changing, not how fast the object is going. At the top the velocity is passing through zero on its way from positive to negative, and passing through zero is exactly what changing rapidly looks like. If the acceleration really were zero there, the velocity would stop changing and the ball would hover.
Worked example
A ball is thrown straight up at 20 metres per second. Find how high it rises and how long until it returns to the thrower's hand. Take up as positive.
Figure (svg): A ball thrown upward shown at five moments with its velocity at each, and a panel noting the acceleration at the top
Set the knowns for the trip up
Why: At the highest point the velocity is zero — that is the extra known the phrase how high supplies.
\[ v_0 = 20, \quad v = 0, \quad a = -9.8 \]
Time is never mentioned, so use the equation without it
Why: This gives the height directly.
\[ 0 = 20^2 + 2(-9.8)\,\Delta x \]
\[ \Delta x = \dfrac{400}{19.6} = 20.4 \text{ m} \]
Now find the time to the top
Why: Use the equation relating the two velocities and the time.
\[ 0 = 20 + (-9.8)t \;\Rightarrow\; t = 2.04 \text{ s} \]
Double it for the round trip
Why: The motion is symmetric: rising from the hand to the top takes exactly as long as falling back.
\[ t_{\text{total}} = 4.08 \text{ s} \]
Verify: by checking the return speed
Why: Putting the full 4.08 seconds into the first equation gives v equals 20 minus 9.8 times 4.08, which is negative 20 metres per second — the same speed as it was thrown, directed downward. That symmetry is a genuine physical fact and makes an excellent check.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 2.7
Socratic
Worth being able to justify rather than assert, since it saves half the work on many problems.
Discussion prompt
Why does the ball take the same time going up as coming down, and return at the same speed?
Hint: What is the rate of change of velocity on the way up compared with on the way down?
Answer:
Because the acceleration is constant and the same throughout, so the velocity changes at a steady rate the whole way.
Going up it loses 9.8 metres per second of speed each second, from 20 down to 0. Coming down it gains at exactly the same rate, so reaching the same height takes the same time and produces the same speed.
The symmetry breaks the moment air resistance matters, because then the force depends on speed and is not the same on the way up as on the way down. That is why problems specify that air resistance is negligible.
Counterexample
The oldest wrong idea in physics, and worth dismantling properly rather than just being told it is wrong.
Discussion prompt
A bowling ball and a tennis ball are dropped together from a first-floor window. Which lands first, and why does the intuition say otherwise?
Hint: Find where mass appears in the four equations.
Answer:
They land together. Mass appears nowhere in any kinematic equation, and in free fall every object has the same acceleration.
The intuition comes from real experience, and the experience is real - but it is about air resistance, not gravity. Drop a feather and a coin and the coin wins, because air affects the feather far more relative to its weight.
Remove the air and the feather and the coin land together, which has been filmed in a vacuum chamber. So the intuition is a correct observation about air, misattributed to gravity.
Section
Where your precalculus pays off
Concept
Everything so far assumed motion along a line. Real motion is rarely along a line, and the moment it is not you need vectors — which is where the trigonometry you finished before this year starts earning its keep.
vector — a quantity with both a size and a direction, such as velocity, displacement or acceleration
The whole strategy for two dimensions is to break every vector into a horizontal piece and a vertical piece, solve two separate one-dimensional problems, and put the answers back together. Nothing new is needed beyond that.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 3.2
Picture it
One vector becomes two numbers, and the two numbers are what you actually compute with.
Figure (svg): A velocity vector of twenty-five metres per second at thirty degrees resolved into horizontal and vertical components using cosine and sine
Cosine takes the adjacent side and sine takes the opposite one, exactly as in your precalculus. The only new thing is remembering which side the angle is measured from — draw the triangle every single time and it cannot go wrong.
Warm-up
You finished precalculus before this year, so this is recall rather than teaching. These five values cover almost every angle a physics problem chooses.
Discussion prompt
Without a calculator: sine and cosine of 30, of 45, and of 60 degrees.
Hint: The two special triangles from your precalculus course.
Answer:
Sine of 30 is exactly one half; cosine of 30 is root three over two, about 0.866.
Sine and cosine of 45 are both root two over two, about 0.707.
Sine of 60 is root three over two, about 0.866; cosine of 60 is exactly one half.
Notice the pattern: 30 and 60 swap their values. Problems pick these angles precisely so the arithmetic stays clean, which is a hint that you are on the intended route when they appear.
Fill the middle
A ball is launched at 25 metres per second at 30 degrees above the horizontal.
Fill in the blanks
v_21.7 m/s = 25\cos 30^\circ = 12.5 m/s, \qquad v____ = 25\sin 30^\circ = ___
Why: Cosine of thirty degrees is root three over two, about 0.866, and 25 times that is 21.7. Sine of thirty is exactly one half, so the vertical component is exactly 12.5. Note that the components are each smaller than the original 25, which they must be — a leg of a right triangle can never beat the hypotenuse.
Elimination
A velocity of 20 metres per second is resolved into components. Three of these are impossible on sight.
Eliminate the wrong options
Which pair of components could be correct?
Survives elimination: v1
Why: Sixteen squared is 256, twelve squared is 144, and those add to 400, whose square root is exactly 20. This is the three-four-five triangle scaled by four, and it is worth recognising because it appears constantly in physics problems chosen to have clean numbers.
Section
Two problems, one clock
Concept
Once launched, and ignoring air resistance, a projectile has gravity acting on it and nothing else. Gravity acts straight down, so it changes the vertical velocity and does absolutely nothing to the horizontal one.
So you solve two separate one-dimensional problems. The only thing linking them is the clock — the time in one is the same as the time in the other, and that shared time is what lets you get from one to the other.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 3.4
Picture it
One ball dropped, one launched sideways at the same instant, photographed at equal time intervals.
Figure (svg): Two balls falling, one dropped straight down and one launched horizontally, shown at five equal time intervals at matching heights
They are level with each other in every frame and hit the ground at the same moment. The sideways motion changes where the second ball is, and not when it lands.
Edge cases
Every model has a range of validity. Worth knowing where this one ends.
Discussion prompt
The whole projectile treatment assumes air resistance is negligible. When does that assumption break, and what changes?
Hint: Which of the two directions loses its constant velocity first?
Answer:
It breaks when the object is light for its size or moving fast - a shuttlecock, a beach ball, a bullet over a long range. Then the drag force is comparable to the weight and cannot be ignored.
What changes is that the horizontal velocity is no longer constant, since drag has a horizontal component that opposes the motion. The path stops being a parabola and becomes lopsided, falling more steeply than it rose.
The range also drops well below the formula's prediction, and the optimum launch angle falls below 45 degrees. AP Physics 1 keeps you inside the no-drag model, but knowing where the edge is stops you trusting it outside the exam.
Check
Solve it on paper before you click.
Check your understanding
A bullet is fired horizontally from a rifle at the same instant an identical bullet is dropped from the same height. Ignoring air resistance and assuming level ground, which lands first?
Answer: C
Why: Landing is entirely a vertical question, and vertically the two bullets are identical: both start with zero vertical velocity and both accelerate downward at 9.8. The horizontal motion of the fired bullet has no influence on how long the fall takes.
Worked example
A ball rolls off a bench 19.6 metres high at 6.0 metres per second horizontally. How long is it in the air, and how far from the bench does it land? Take down as positive for the vertical part.
Figure (svg): Two balls falling, one dropped and one launched horizontally, level with each other at every time interval
Split into two problems
Why: Vertically it is a drop; horizontally it is constant velocity. Solve the vertical one first, because it is the one that gives you the time.
Vertical: the initial vertical velocity is zero
Why: Rolling off means it leaves horizontally, with no vertical velocity at all — that is the given people miss.
\[ 19.6 = 0 + \tfrac{1}{2}(9.8)t^2 \;\Rightarrow\; t^2 = 4.0 \]
\[ t = 2.0 \text{ s} \]
Horizontal: no acceleration, so distance is velocity times time
Why: Use the time you just found, because both directions share the same clock.
\[ x = v_x t = (6.0)(2.0) = 12 \text{ m} \]
Verify: that the horizontal velocity never changed
Why: The ball lands still moving at 6.0 metres per second horizontally, along with a downward velocity of 9.8 times 2.0, which is 19.6. If your working ever changes the horizontal velocity, something acted sideways, and in this problem nothing did.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 3.4
Worked example
A ball is launched from ground level at 25 metres per second at 30 degrees above the horizontal. Find its time of flight, its range on level ground, and its maximum height. Take up as positive.
Figure (svg): A parabolic projectile path plotted to a range of about fifty-five metres and a peak near eight metres, with the top marked
Resolve the launch velocity
Why: This is the step where the trigonometry happens, and after it there is no more trigonometry in the problem.
\[ v_{0x} = 21.7, \qquad v_{0y} = 12.5 \]
Vertical: find the time to the top
Why: At the top the vertical velocity is zero.
\[ 0 = 12.5 - 9.8t \;\Rightarrow\; t = 1.28 \text{ s} \]
Double it for the whole flight
Why: Level ground, so the rise and the fall take equal times.
\[ T = 2.55 \text{ s} \]
Horizontal: range is the constant velocity times the total time
Why: The horizontal velocity never changed, so this is a single multiplication.
\[ R = (21.7)(2.55) = 55.2 \text{ m} \]
Maximum height, from the vertical motion alone
Why: Use the equation without time, on the trip up.
\[ H = \dfrac{12.5^2}{2(9.8)} = 7.97 \text{ m} \]
Verify: the range with the standard formula
Why: For level ground the range is v0 squared times the sine of twice the angle, all over g. That gives 625 times the sine of 60, which is 541.3, divided by 9.8, which is 55.2 metres. It agrees with the componentwise answer, so both the trigonometry and the timing are right.
HyperPhysics, Mechanics — motion in one and two dimensions projectile motion
Comparison
Fill the blank from the pattern rather than from memory. This table is the whole of projectile motion.
Comparison matrix
| question | horizontal | vertical |
|---|---|---|
| what is the acceleration | zero | 9.8 downward |
| does the velocity change | no, it is constant throughout | yes, it changes by 9.8 every second |
| which equation applies | distance equals velocity times time | all four kinematic equations |
| what happens at the top | nothing, it carries on unchanged | the velocity is momentarily zero |
| what links it to the other column | the time | the time |
The last row is the one to hold on to. The two columns are independent in every respect except that they happen simultaneously, and the shared time is the bridge you cross to get from one to the other.
Ranking
Doing these out of order is what makes projectile problems feel hard.
Put in order
Why: Draw before anything, so the signs are settled. Resolve next, because every later step needs the components. Then vertical before horizontal — always — because the vertical direction is the one with acceleration in it and therefore the one that determines the time. The horizontal direction cannot give you a time on its own, since it has two unknowns in one equation.
Missing information
Under-specified problems appear on homework, sometimes deliberately.
Discussion prompt
A ball is launched at 30 metres per second and lands 60 metres away. Find the launch angle. What has been assumed that was not stated?
Hint: Which formula were you about to use, and what does it quietly require?
Answer:
That the launch and landing are at the same height. The range formula only holds on level ground, and if the ball was thrown from a cliff or into a basket the answer changes completely.
Also that air resistance is negligible, which every problem at this level assumes but not every problem says.
With those assumptions the range formula gives sine of twice the angle equals 60 times 9.8 over 900, which is 0.653, so twice the angle is about 40.8 degrees and the angle is about 20.4 degrees. There is also a second valid answer near 69.6 degrees, since two angles give the same range — and noticing that second solution is worth a mark on its own.
Section
Relative motion
Concept
A boat crossing a flowing river, or a plane in a crosswind, is moving relative to something that is itself moving. The velocities add, and they add as vectors rather than as numbers.
This is the last new idea in the unit and it is a short one, because you already know how to add vectors: break both into components, add the components separately, and reassemble.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 3.5
Picture it
The standard picture, with numbers chosen so the arithmetic is clean.
Figure (svg): A boat velocity of four metres per second across a river combined with a three metres per second current, giving a five metres per second resultant
The resultant is 5.0, not 7.0, because the two velocities are perpendicular. And the crossing time is unchanged by the current — the current moves the boat downstream, not across, so the across-the-river problem is untouched by it.
Two truths and a lie
The boat points straight across at 4.0 metres per second; the current runs at 3.0.
Eliminate the wrong options
Cross out the false statement.
Survives elimination: r3
Why: Crossing is entirely an across-the-river question, and the current has no across-the-river component at all. The time to cross depends only on the width and the 4.0 metres per second, exactly as it would in still water. The current changes where you land, not when you arrive — which is the same independence idea as the projectile, in a different costume.
Section
Consolidation
Pattern
Everything above, compressed to what you would want on a single index card.
| situation | what you know immediately | what to do first |
|---|---|---|
| starts from rest | initial velocity is zero | count your knowns, you have three |
| comes to a stop | final velocity is zero | same |
| dropped | initial velocity zero, a is 9.8 down | solve vertically |
| thrown straight up | at the top v is zero, a is unchanged | split into up and down, or use symmetry |
| rolls off a ledge | initial vertical velocity is zero | vertical first, to get the time |
| launched at an angle | resolve into components | vertical first, to get the time |
| two velocities at once | add them as vectors | components, then reassemble |
Notice how many rows say vertical first. When there is acceleration in one direction and not the other, the accelerated direction is the one that sets the clock.
College Board, AP Physics 1: Algebra-Based Course and Exam Description Unit 1
Check
Solve it on paper before you click.
Check your understanding
A stone is thrown horizontally at 12 metres per second from a cliff and lands 45 metres from the base. You are asked for the height of the cliff. What is the first calculation?
Answer: B
Why: This one reverses the usual order, and that is what makes it a good question. Vertically you know only the initial velocity and the acceleration — two knowns, not enough. Horizontally you know the velocity and the distance, so the time comes out as 45 divided by 12, which is 3.75 seconds. That time then unlocks the vertical direction.
Error analysis
A ball is dropped from rest and falls for 3.0 seconds. A student computes the distance fallen.
Annotate
On: \( \Delta x = v_0 t + \tfrac{1}{2}at^2 = (9.8)(3.0) + \tfrac{1}{2}(9.8)(3.0)^2 \)
This is exactly what step two of the method prevents. Writing v0 equals 0 and a equals 9.8 on separate lines before touching an equation makes this mistake almost impossible.
Concept
Before an answer goes on the page it should survive two questions, and both take about five seconds.
Nearly every catastrophic error - a factor of ten, a missing square, a swapped sine and cosine - shows up in one of those two checks. Neither requires you to know whether the answer is right, only whether it is possible.
OpenStax, College Physics 2e — chapters 2 and 3 chapter 1.3
Explain it
The real test of whether today landed. Say it out loud, to a person or to an empty room.
Discussion prompt
In under a minute, teach someone how to pick the right kinematic equation without guessing.
Hint: Five quantities, three given, one wanted, one left over.
Answer:
List the five quantities: initial velocity, final velocity, acceleration, time, displacement.
Mark the three the problem gives you and the one it asks for. That leaves exactly one quantity that is neither given nor wanted.
Use the equation that does not contain that quantity. There is exactly one, because each of the four equations omits a different one - so the choice is forced rather than guessed.
Explain it to yourself
If you can say this without notes, the hardest part of the unit has landed.
Discussion prompt
In two sentences, why can a projectile problem be split into two separate problems?
Hint: Which direction does the only force point, and what does that leave untouched?
Answer:
Because gravity acts only downward, so it changes the vertical velocity and leaves the horizontal one completely alone.
The two directions therefore evolve independently, and the only thing they share is the elapsed time — which is what lets you solve one direction and carry the time across to the other.
Real world
Worth grounding, because the unit can feel like invented problems about balls.
Discussion prompt
Name three everyday situations that are exactly the problems in this unit, and say which one each is.
Hint: Think of one with braking, one with an arc, and one with two motions at once.
Answer:
Braking distance on a road sign is the equation without time — an initial speed, a final speed of zero, and a deceleration.
A basketball shot is an angled projectile, and the reason a high arc is more forgiving is that the ball approaches the hoop closer to vertical, making the effective target larger.
An aeroplane correcting for a crosswind is the river problem: the pilot points partly into the wind so that the resultant velocity is along the intended track.
Warm-up
Close the deck. This is worth several times what rereading is worth, and it tells you exactly what to revise.
Discussion prompt
From memory: the five quantities; the rule that decides speeding up from slowing down; how to choose an equation in one step; the acceleration at the top of a throw; the two components of a launch at 25 metres per second and 30 degrees; the horizontal acceleration of a projectile; what links the two directions; and the resultant of 4.0 across and 3.0 downstream.
Hint: Five quantities, one sign rule, one trick, one number, two components, one zero, one link, one hypotenuse.
Answer:
Initial velocity, final velocity, acceleration, time and displacement.
Matching signs means speeding up; opposite signs means slowing down.
Find the quantity the problem never mentions, and use the equation that omits it.
9.8 metres per second squared downward — unchanged, even though the velocity is zero.
About 21.7 horizontally and exactly 12.5 vertically.
Zero. The horizontal velocity is constant for the entire flight.
The time. Nothing else.
5.0 metres per second, from the three-four-five triangle.
Connect it up
Do this before the next session and bring it. It is the single most useful thing you can make in this unit.
Draw it
On one side, write the four equations and mark beside each the quantity it omits. On the other side, draw three pictures: a ball thrown straight up with the velocity and acceleration marked at three moments, a horizontal launch off a ledge, and an angled launch with its components labelled. Add nothing else.
If it fits on a card, you understand it. If it needs a second card, the extra material is the part to work on.
Exit ticket
Your school is still on the maths refresher, so we can spend the next hour wherever it helps most.
Predict first
What would be most useful next time?
Correct: Whichever you pick is what I will prepare.
Why: Working through the real homework is usually the strongest choice once the method is in place, because it applies everything under the conditions that actually matter. Worth knowing that motion graphs are a whole topic of their own and there is a separate deck for them, so that option is a full session rather than a detour.
Recap
One unit, one method. Everything here is the same five steps applied to a slightly harder picture each time.
| equation | what it omits | the phrase that signals it |
|---|---|---|
| v = v0 + a t | displacement | how fast after this long |
| dx = v0 t + half a t squared | final velocity | how far in this long |
| v squared = v0 squared + 2 a dx | time | how fast after this far |
| dx = half (v0 + v) t | acceleration | both speeds are given |
| x = vx t | nothing, no acceleration | the horizontal part of a projectile |
College Board, AP Physics 1: Algebra-Based Course and Exam Description Unit 1 — with OpenStax chapters 2 and 3 for extra worked practice
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