Unit 1 of an algebra and trigonometry based honors physics course, built specifically around motion graphs rather than around formulas. It fixes the vocabulary first - distance against displacement, speed against velocity, and the sign rule that decides speeding up from slowing down - then teaches position-time graphs as height plus slope, velocity-time graphs as height plus slope plus area, and finally the conversion between them by accumulating area. The constant-acceleration equations arrive last and are presented as graph areas already solved, with every worked answer cross-checked against the geometry. No calculus is used anywhere.
Subject: Honors Physics · 61 slides · diagram-first lesson
Open the interactive version of this deck · Homework for this lesson
Title
Honors Physics - Unit 1
Reading motion off a graph, and turning one graph into another
Objectives
You said graphs are the sticking point, and specifically going from a velocity graph to a position graph. That is the destination of this deck, and everything before it exists to make that conversion obvious rather than memorised.
Two sentences do most of the work in this unit, and they are worth learning before anything else.
James S. Walker, Physics — the course text (kinematics in one dimension) Ch. 2 — The chapter this unit follows.
Concept
Everything in this deck is one of these two ideas wearing different clothes. Learn them now and the rest is bookkeeping.
\[ \textbf{slope of a graph} \;=\; \text{the rate at which the plotted quantity is changing} \]
\[ \textbf{area under a graph} \;=\; \text{the total accumulated change in the quantity above it} \]
On a position graph, the slope is velocity. On a velocity graph, the slope is acceleration and the area is displacement. That is the whole ladder, and it is the same ladder every time.
Section
Half the lost marks in this unit are vocabulary
Warm-up
Answer before reading on. Most people get one of the two numbers wrong.
Discussion prompt
You walk 8 metres east, then turn round and walk 3 metres west. What distance did you travel, and what was your displacement?
Hint: One of these cares about the path. The other only cares about the endpoints.
Answer:
\[ \text{distance} = 8 + 3 = 11 \text{ m}, \qquad \text{displacement} = +8 - 3 = +5 \text{ m} \]
Distance adds up every step regardless of direction, so it can never decrease. Displacement only looks at where you started and where you ended, so walking backwards subtracts. Displacement can be negative, and it can be zero even after a long walk.
Concept
These three get used interchangeably in ordinary speech and they are not interchangeable here.
Position — Where the object is, measured from an origin you choose. It carries a sign, so a position can be negative.
Distance — The total length of path travelled. Always positive, never decreases, ignores direction entirely.
Displacement — The change in position: final minus initial. Carries a sign, and only the two endpoints matter.
\[ \Delta x = x_{\text{final}} - x_{\text{initial}} \]
The delta symbol always means final minus initial, in that order. Getting the order backwards flips every sign in the problem, and it is the single most common arithmetic slip in this unit.
Picture it
Drawn on a line, the difference stops being abstract.
Figure (svg): A number line from zero to ten with an arrow running from zero to eight and a shorter arrow returning from eight to five, showing a path length of eleven metres but a net displacement of five metres.
When a question says how far did it travel, it wants distance. When it says how far from the start, or asks for a change, it wants displacement.
Concept
The same split runs through the next pair, and for the same reason: one of them keeps the direction and one throws it away.
\[ \text{average speed} = \frac{\text{distance travelled}}{\text{time}} \qquad \text{average velocity} = \frac{\Delta x}{\Delta t} \]
Speed uses distance, so it is always positive. Velocity uses displacement, so it carries a sign, and the sign is the direction.
For the walk on the last slide taken over 10 seconds, the average speed is 1.1 metres per second and the average velocity is 0.5 metres per second. Both are correct. They are answers to different questions.
Discrimination
Real homework wording. Sort by what the question actually wants.
Sort into buckets
Sort each phrase by the quantity it is asking for.
Concept
Acceleration is the third rung of the same ladder, and it is where intuition starts betraying people.
\[ a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t} \]
It measures how fast the velocity is changing, not how fast the object is going. An object can be moving very fast with zero acceleration, and it can be momentarily at rest while accelerating hard.
A ball at the very top of its flight is the standard example. Its velocity is zero for an instant. Its acceleration is not zero at all, which is precisely why it does not stay up there.
Trap
The acceleration is negative, so the object must be slowing down. Negative means less, and less speed means slowing.
\[ a = -3 \text{ m/s}^2 \;\Longrightarrow\; \text{slowing down} \]
The sign of the acceleration alone tells you nothing about speeding up or slowing down. What decides it is whether the acceleration and the velocity have the same sign or opposite signs.
| velocity | acceleration | what happens |
|---|---|---|
| positive | positive | speeding up, moving forward |
| positive | negative | slowing down, moving forward |
| negative | negative | speeding up, moving backward |
| negative | positive | slowing down, moving backward |
Rows two and three are the ones people miss. A ball thrown upward and a ball falling back down have exactly the same negative acceleration throughout, yet one is slowing and the other is speeding up.
Say it as a rule: same signs means speeding up, opposite signs means slowing down. Never read the sign of the acceleration on its own.
Definition probe
Use the rule rather than intuition. Intuition is what the trap was built out of.
Sort into buckets
Sort each case by what the object is doing.
Concept
One more split, and it is the one that makes the graphs necessary rather than optional.
Average velocity — Displacement divided by the whole time interval. It compares two moments and ignores everything in between.
Instantaneous velocity — The velocity at one single moment, which is what a speedometer shows.
A car that drives 60 kilometres in one hour has an average speed of 60 kilometres per hour even if it stopped twice and hit 100 in between. The average hides the story, and the graph is where the story lives.
Two truths and a lie
Four statements. Exactly one survives contact with the definitions.
Eliminate the wrong options
Rule out the three that misuse a definition, and keep the one that is true.
Survives elimination: v1
Why: If you travelled no path at all then you never moved, so you certainly ended where you started and the displacement is zero. The reverse does not hold, because you can travel a long path and return. Distance zero forces displacement zero; displacement zero forces nothing.
Real world
A running app reports both a distance and a straight-line map view of your route.
Discussion prompt
You run a loop that finishes at your front door. What does the app's distance figure say, and what would your displacement be?
Hint: Which number can a loop make zero?
Answer:
The distance is the full length of the loop, perhaps five kilometres, because it accumulates every step regardless of direction.
The displacement is exactly zero, because you finished at the point you started from. Your average velocity for the run is also exactly zero, while your average speed was perfectly respectable.
This is why nobody reports average velocity for a run. For loops it is always zero and it tells you nothing, whereas average speed tells you how the run actually went.
Section
The height is where, the slope is how fast
Concept
A position-time graph puts time along the bottom and position up the side. Two separate pieces of information live in it, and beginners tend to read only the first.
\[ v = \text{slope} = \frac{\Delta x}{\Delta t} = \frac{\text{rise}}{\text{run}} \]
A line going up means moving in the positive direction. A line going down means moving in the negative direction. A flat line means not moving at all, which is the reading people most often get wrong.
Picture it
Read each segment before looking at the labels.
Figure (svg): A position-time graph with three straight segments: a rising line from zero to forty metres over four seconds, a horizontal line at forty metres for three seconds, and a falling line back to ten metres over five seconds.
The flat middle section is the one to stare at. The object is far from the origin and completely motionless. A big height with zero slope is a common source of confusion.
Worked example
Reading the three velocities from the graph on the previous slide, one segment at a time.
Take the first segment and compute its slope as rise over run.
Why: Velocity on a position graph is the slope, so this is not a formula being applied to the graph, it is the definition of what the graph's steepness means.
\[ v_1 = \frac{40 - 0}{4 - 0} = \frac{40}{4} = +10 \text{ m/s} \]
Read the middle segment.
Why: The height does not change, so the rise is zero, so the slope is zero. The object sits at 40 metres without moving.
\[ v_2 = \frac{40 - 40}{7 - 4} = 0 \text{ m/s} \]
Take the third segment, keeping the sign.
Why: The line falls, so the rise is negative and the velocity is negative. That negative sign is the physics: the object is heading back toward the origin.
\[ v_3 = \frac{10 - 40}{12 - 7} = \frac{-30}{5} = -6 \text{ m/s} \]
Verify: check the total displacement two different ways.
Why: From the graph endpoints, the displacement is 10 minus 0, which is plus 10 metres. Adding the segments: plus 40, then 0, then minus 30, which is also plus 10 metres. The two methods agree, so the slopes were read correctly.
Prediction
A position-time graph that curves upward, getting steeper as time goes on.
Predict first
What is the object doing?
Correct: Moving forward and speeding up
Why: Steepness is velocity, so a graph getting steeper means the velocity is increasing. The line still rises, so the motion is forward. Curvature on a position graph is exactly what acceleration looks like: a straight line means constant velocity and zero acceleration, while any bend means the velocity is changing.
Picture it
Compare the two shapes directly. The difference is the whole idea of acceleration, seen rather than defined.
Figure (svg): Two position-time curves on one set of axes: a straight line rising steadily and a curve that starts shallow and becomes progressively steeper.
On a position graph, acceleration is curvature. That is the whole visual signature, and it is why a straight position graph can never show acceleration.
Socratic
Homework often gives you an unlabelled sketch and asks a comparison question. Those are not harder, they are just missing the crutch.
Discussion prompt
Two objects are drawn on the same position-time axes. Their lines cross at one point. What is physically happening at that moment, and what is definitely not happening?
Hint: The crossing point is where two heights are equal. What is height on this graph?
Answer:
What is happening: the two objects are at the same position at the same time. On a road, that means they are side by side. Same height, same moment.
What is not happening: they are not travelling at the same speed. Equal speeds would mean equal slopes, which on a graph looks like parallel lines, not crossing ones. In fact crossing guarantees the slopes are different at that point.
Same position is a crossing. Same velocity is parallel. Keeping those two apart handles most of the unlabelled graph questions you will be asked.
Ranking
Four straight segments on the same position-time axes. Speed is steepness, ignoring the direction of the tilt.
Put in order
Why: Speeds are 0, then 2, then 6, then 10 metres per second. The falling line is third because speed ignores the sign: its velocity is minus 6 but its speed is 6. Sorting by velocity instead would put the falling line first, which is exactly the distinction being tested.
Faded example
A position graph passes through 12 metres at 2 seconds and 45 metres at 7 seconds.
Fill in the blanks
v = \frac45 - 126.6 = \frac___}___ = ___ \text___
Why: The rise is the change in position, final minus initial, which is 45 minus 12 or 33 metres. The run is 5 seconds. Dividing gives 6.6 metres per second. Writing the subtraction as final minus initial rather than the other way round is what keeps the sign correct when the graph falls instead of rising.
Counterexample
A student draws a position-time graph with a perfectly vertical segment.
Discussion prompt
What would a vertical segment on a position-time graph mean physically, and why can it never happen?
Hint: Work out the slope of a vertical line.
Answer:
A vertical segment means the position changes by some amount while no time passes at all. The slope is a division by zero, so the velocity would be infinite.
That is the object being in two places at once, which nothing does. So a position-time graph can be steep, but it can never be vertical, and a graph drawn that way has an error in it rather than an exotic motion.
The same reasoning says a position graph can never double back to the left, because that would put the object in two places at the same time. Time only ever runs forward across the page.
Section
Now the slope means something new, and the area means something at all
Concept
Swap the vertical axis from position to velocity and both readings shift up the ladder. This is the step that confuses people, because the graph looks the same and means something different.
| on a position graph | on a velocity graph |
|---|---|
| height is position | height is velocity |
| slope is velocity | slope is acceleration |
| area means nothing useful | area is displacement |
The third row is new. Area under a position graph has no physical meaning, which is why nobody ever asks for it. Area under a velocity graph is displacement, and questions ask for it constantly.
Notation
This is worth seeing once as arithmetic rather than accepting as a rule.
Annotate
On: \( \text{area} = v \times \Delta t = \left(\frac{\Delta x}{\Delta t}\right) \times \Delta t = \Delta x \)
Rectangles, triangles and trapeziums are all the geometry this unit needs. No calculus is required to find any area you will be asked for.
Worked example
A car accelerates from rest to 20 metres per second over 8 seconds, then holds that speed for 4 more seconds.
Split the region under the graph into shapes you can compute.
Why: Breaking an awkward region into a triangle plus a rectangle is the standard move, and it is always available for the straight-line graphs in this unit.
Find the area of the triangle covering the acceleration phase.
Why: The velocity climbs from 0 to 20 over 8 seconds, so the region is a triangle with base 8 and height 20.
\[ A_1 = \tfrac{1}{2} \times 8 \times 20 = 80 \text{ m} \]
Find the area of the rectangle covering the constant-speed phase.
Why: The velocity is a flat 20 for 4 seconds, so the region is a rectangle.
\[ A_2 = 20 \times 4 = 80 \text{ m} \]
Add the pieces to get the total displacement.
Why: Both regions are above the time axis, so both count as positive displacement and they simply add.
\[ \Delta x = A_1 + A_2 = 80 + 80 = 160 \text{ m} \]
Verify: check the first phase with the average-velocity shortcut.
Why: When acceleration is constant, the average velocity is the average of the start and end values: half of 0 plus 20, which is 10 metres per second. Over 8 seconds that gives 80 metres, matching the triangle exactly. Two methods, one answer.
Figure (svg): A velocity-time graph rising from zero to twenty metres per second over eight seconds then flat for four more, with the area beneath shaded and split into a triangle and a rectangle.
Estimation
A velocity-time graph spends 4 seconds at plus 5 metres per second, then 4 seconds at minus 5 metres per second.
Predict first
What is the total displacement over the 8 seconds?
Correct: 0 metres
Why: The first region is a rectangle of area plus 20 metres. The second is the same size but sits below the time axis, so it counts as minus 20 metres. They cancel exactly, so the object finishes where it started. The distance travelled is 40 metres, which is a different question: distance adds the sizes of the areas and ignores their signs.
Trap
The graph dips below the time axis between 6 and 10 seconds. Areas cannot be negative, so that region is added to the total like any other, giving a displacement of 55 metres.
Geometric area is never negative, but the area under a velocity graph is not being used as geometry. It is being used as a displacement, and displacement has a direction.
A velocity below the axis means the object is moving in the negative direction, so during that interval it is undoing displacement it built up earlier. The region has to be subtracted.
\[ \Delta x = A_{\text{above}} - A_{\text{below}} \]
This is exactly the distance versus displacement split from Part 1, arriving in graph form. If the question asks how far it travelled, add the sizes. If it asks for displacement or final position, subtract the parts below the axis.
Concept
The other reading on a velocity graph is its steepness, and one rung up the ladder that means acceleration.
\[ a = \text{slope} = \frac{\Delta v}{\Delta t} \]
Notice that a flat velocity graph high above the axis means fast and steady. Flat does not mean stopped here. Flat means stopped only on a position graph.
Matching
The point of this exercise is that graph shape alone is not enough. You must check which axis you are on.
Match the pairs
Why: The two flat lines mean completely different things: on a position graph nothing is moving, while on a velocity graph the object may be moving very fast and simply not changing speed. The two steep lines split the same way, into fast motion on one graph and hard acceleration on the other. Always read the vertical axis label before interpreting a shape.
Pattern
This is the single organising idea of the unit. Going down the ladder you take slopes. Going up the ladder you take areas.
| from | operation | to |
|---|---|---|
| position graph | take the slope | velocity |
| velocity graph | take the slope | acceleration |
| acceleration graph | take the area | change in velocity |
| velocity graph | take the area | change in position |
Slopes go down the ladder, areas go back up it. Once you can say which direction a question is asking you to move, the method is already chosen for you.
Check
Solve it on paper before you click.
Check your understanding
A velocity-time graph is a horizontal line at 15 metres per second from 0 to 6 seconds. What are the acceleration and the displacement over that interval?
Answer: B
Why: The line is flat, so its slope is zero and the acceleration is zero. The region beneath it is a rectangle of height 15 and width 6, so the displacement is 90 metres. Flat on a velocity graph means steady motion, not no motion.
Invariant
A velocity graph held flat at 6 metres per second. Watch the running total, because that total is the position graph being drawn.
Step through it
What is happening to the size of each new strip of area, and what does that do to the position graph?
Now imagine the velocity climbing instead. Each strip would be taller than the last, the position would gain more each second, and the position graph would curve upward. That is the entire conversion in one sentence.
Anomaly
Object A holds a steady 10 metres per second for 6 seconds. Object B accelerates uniformly from 0 to 20 metres per second over the same 6 seconds.
Predict first
How do their displacements compare?
Correct: They are identical
Why: A gives a rectangle of area 10 times 6, which is 60 metres. B gives a triangle of area half times 6 times 20, which is also 60 metres. Different shapes, identical areas, identical displacement. This is worth sitting with, because it shows that the shape of the motion and the amount of the motion are genuinely separate questions.
Edge cases
Push the idea to its edge, which is usually where the definition becomes clear.
Discussion prompt
A velocity graph touches 8 metres per second at exactly one instant and is zero elsewhere. What displacement does that instant contribute?
Hint: What is the width of the region under a single instant?
Answer:
None at all. The region under a single instant has zero width, so its area is zero however tall it is.
This is why displacement always depends on an interval rather than a moment. Velocity is a property of an instant; displacement is a property of a stretch of time.
It is also the reason the ball at the top of its flight goes nowhere during that instant despite everything else that is happening to it.
Section
The thing you actually asked for
Concept
Going from a velocity graph to a position graph is a single question asked repeatedly: what is the area so far?
A useful check before you draw anything: a velocity graph above the axis must always produce a position graph that rises, because positive velocity means the position is increasing, no matter what the velocity is doing otherwise.
Picture it
Read the top graph left to right and watch what each feature does to the bottom one.
Figure (svg): Two vertically aligned graphs sharing a time axis: on top a velocity graph rising from zero to twenty then holding flat, and beneath it the matching position graph curving upward then becoming a straight rising line.
The curve on the bottom is not decoration. It is there because the slope of the position graph has to keep increasing while the velocity above it is still climbing.
Worked example
The car from earlier: 0 to 20 metres per second over 8 seconds, then steady at 20 for 4 more. Starting position is 0.
Mark the starting height on the position graph.
Why: The position graph needs a starting value, and the velocity graph cannot supply it. It comes from the problem, and here the car starts at the origin.
Work out the accumulated area at each landmark time.
Why: These become the heights of the position graph. Landmarks are wherever the velocity graph changes character.
\[ x(8) = \tfrac{1}{2} \times 8 \times 20 = 80 \text{ m}, \qquad x(12) = 80 + (20 \times 4) = 160 \text{ m} \]
Decide the shape between the landmarks, not just the values.
Why: From 0 to 8 seconds the velocity is increasing, so the slope of the position graph increases: the curve bends upward. From 8 to 12 the velocity is constant, so the position graph is a straight line.
Check the slope of the straight section matches the velocity.
Why: The final section runs from 80 metres to 160 metres over 4 seconds, giving a slope of 20 metres per second, which is exactly the height of the velocity graph over that interval.
Verify: test one interior point on the curve.
Why: At 4 seconds the velocity has reached 10 metres per second, so the accumulated area is half of 4 times 10, which is 20 metres. The curve should pass through 20 metres at 4 seconds, well below the 40 metres a straight line to the point at 8 seconds would predict. That gap is the curvature, confirmed numerically.
Figure (svg): The completed position-time graph curving upward from the origin to eighty metres at eight seconds, then continuing as a straight line to one hundred and sixty metres at twelve seconds.
Fill the middle
A different velocity graph: constant at 6 metres per second for 5 seconds, then constant at 2 metres per second for 5 more. Position starts at 10 metres.
Fill in the blanks
x(0) = 10, \qquad x(5) = 40, \qquad x(10) = 50
Why: The first area is 6 times 5, which is 30 metres, added to the starting 10 gives 40 metres. The second area is 2 times 5, which is 10 metres, giving 50 metres. Both sections are straight on the position graph because both velocities are constant, but the second is much shallower because the velocity dropped.
Prediction
A velocity graph starts at plus 8 metres per second and falls steadily, crossing zero at 4 seconds and continuing down to minus 8 at 8 seconds.
Predict first
What does the matching position graph do?
Correct: Rises, levels off at 4 seconds, then falls back
Why: While the velocity is positive the position increases, but ever more slowly as the velocity shrinks, so the graph rises and flattens. At 4 seconds the velocity is zero, which is the peak of the position graph. After that the velocity is negative so the position decreases. The last option confuses the total displacement over the whole interval, which is indeed zero, with the shape of the graph in between, which is certainly not flat.
Picture it
This shape is worth recognising instantly, because it is the shape of anything thrown straight up.
Figure (svg): Two aligned graphs: a velocity graph falling steadily from plus eight through zero at four seconds down to minus eight, and beneath it a position graph rising to a peak at four seconds and coming back down.
Where the velocity graph crosses the axis, the position graph turns round. That correspondence is worth memorising because it appears on almost every test in this unit.
Explain it to yourself
The moment a thrown ball is at its highest point.
Discussion prompt
At the top of the flight the velocity is zero. Why is the acceleration not zero at that same instant?
Hint: Ask what would happen next if the acceleration really were zero there.
Answer:
Velocity being zero is a statement about that single instant. Acceleration is a statement about how the velocity is changing across that instant, and the velocity is changing very much: it goes from slightly positive just before, to slightly negative just after.
If the acceleration truly were zero at the top, the velocity would have no reason to change, so the ball would stay at zero velocity and hover there forever. It does not, so the acceleration is not zero.
On the graph this is easy to see: the velocity line crosses the axis with a definite tilt. It is the tilt that is the acceleration, not the height, and a line can cross zero while remaining just as steep as it was.
Worked example
The reverse conversion is easier, because it only needs slopes rather than accumulated areas. Use the three-segment graph from Part 2.
Split the position graph at every point where its shape changes.
Why: Each straight section has one constant slope, so each becomes one flat section on the velocity graph. The corners are where the velocity jumps.
Compute the slope of each section and plot it as a height.
Why: The slope of a position section is the velocity during it, so a number that was a steepness becomes a height on the new graph.
\[ v_1 = +10, \qquad v_2 = 0, \qquad v_3 = -6 \quad \text{(m/s)} \]
Draw each as a horizontal line over the matching time interval.
Why: Constant velocities are flat lines. The third one sits below the time axis because it is negative, which is how the velocity graph records the return journey.
Verify: check the areas reproduce the original position changes.
Why: The first rectangle is 10 times 4, which is plus 40 metres, matching the climb from 0 to 40. The third is minus 6 times 5, which is minus 30 metres, matching the fall from 40 to 10. Converting back recovers the graph we started from, so the conversion is right.
Figure (svg): A velocity-time graph made of three flat segments: plus ten metres per second for four seconds, zero for three seconds, then minus six metres per second for five seconds below the axis.
Reverse engineer
A position graph is a straight line passing through 5 metres at 1 second and 29 metres at 7 seconds.
Fill in the blanks
v = \frac4___ = ___ \text___
Why: The rise is 24 metres over a run of 6 seconds, giving 4 metres per second. Because the position graph is straight, this velocity holds for the whole interval, so the matching velocity graph is a single flat line at height 4.
Explain it
A classmate converts a velocity graph that is flat at 5 metres per second for 4 seconds into a position graph.
\[ \text{classmate draws: a flat line at } x = 5 \text{ for four seconds} \]
Discussion prompt
What did they get wrong, and what should the position graph look like instead?
Hint: They copied the height across. What were they supposed to do with it?
Answer:
They carried the height of the velocity graph straight onto the position graph. That is the single most common conversion error, and it produces a graph saying the object sat motionless at 5 metres.
The height of the velocity graph is a slope on the position graph, not a height. A constant 5 metres per second means the position graph rises steadily with slope 5, reaching 20 metres after 4 seconds.
A quick sanity check catches this instantly: a positive velocity must produce a rising position graph. Their graph was flat, so the object was not moving, which contradicts the velocity they started from.
Section
Same physics, algebra rather than geometry
Concept
When the acceleration is constant, the graph areas can be worked out once and for all and written as formulas. These are not new physics, they are the same slopes and areas already done for you.
\[ v = v_0 + a t \]
\[ x = x_0 + v_0 t + \tfrac{1}{2} a t^2 \]
\[ v^2 = v_0^2 + 2a(x - x_0) \]
Every one of them assumes the acceleration does not change. Applied to a problem where it does, they give confident and completely wrong answers, so check that assumption before reaching for them.
Comparison
Choosing an equation is a matter of noticing which quantity the problem never mentions.
Comparison matrix
| The quantity not mentioned | Reach for |
|---|---|
| displacement is not involved | v = v0 + at |
| final velocity is not involved | the equation with one half a t squared |
| time is not involved | v squared = v0 squared + 2a times displacement |
Read the problem, list what you have and what you want, and the missing quantity picks the equation. That is faster and far more reliable than trying to remember which formula is for which situation.
Worked example
A car travelling at 20 metres per second brakes at a constant 4 metres per second squared until it stops. How far does it travel?
Note the sign convention before touching any numbers.
Why: The car moves in the positive direction and the braking opposes that motion, so the acceleration is negative. Writing minus 4 rather than 4 here is what keeps every later sign honest.
Choose the equation using the quantity the problem never mentions.
Why: The problem gives an initial velocity, a final velocity of zero and an acceleration, and asks for a distance. Time is never mentioned, which selects the third equation.
\[ v^2 = v_0^2 + 2a(x - x_0) \]
Substitute and solve for the displacement.
Why: The final velocity is zero because the car stops, which is the piece of information hidden inside the word stops.
\[ 0 = 20^2 + 2(-4)(\Delta x) \;\Longrightarrow\; \Delta x = \frac{400}{8} = 50 \text{ m} \]
Verify: redo it as a triangle on a velocity graph.
Why: Stopping from 20 metres per second at 4 metres per second squared takes 5 seconds. The velocity graph is a triangle with base 5 and height 20, so the area is half of 5 times 20, which is 50 metres. The geometry and the algebra agree, which is the point of this whole deck.
Figure (svg): A velocity-time graph falling in a straight line from twenty metres per second at time zero to zero at five seconds, with the triangular region beneath it shaded and labelled fifty metres.
Error analysis
The same braking problem, worked by a student. One line is wrong and it is not the arithmetic.
Annotate
On: \( 0 = 20^2 + 2(4)(\Delta x) \;\Longrightarrow\; \Delta x = -50 \text{ m} \)
Set the sign convention before substituting anything, and then sanity-check whether the answer's sign makes physical sense. Those two habits catch nearly every sign error in this unit.
Check
Solve it on paper before you click.
Check your understanding
An object starts from rest and accelerates uniformly at 3 metres per second squared for 6 seconds. How far does it travel, and what is its final velocity?
Answer: B
Why: The final velocity is the initial velocity plus acceleration times time, which is 0 plus 3 times 6, giving 18 metres per second. The displacement is half times 3 times 6 squared, which is 54 metres. Checking with the graph: a triangle of base 6 and height 18 has area half times 6 times 18, which is also 54 metres.
Concept
The next thing your course does with these equations is drop things. Nothing new is needed, because free fall is the constant-acceleration case with the acceleration already known.
\[ a = -g = -9.8 \text{ m/s}^2 \quad \text{taking upward as positive} \]
The minus sign is a consequence of choosing upward as positive, not a property of gravity. If you choose downward as positive then the acceleration is plus 9.8, and every answer still comes out right as long as you stay consistent.
Everything from this deck transfers directly. A ball thrown upward has a velocity graph that is a straight line of slope minus 9.8, crossing zero at the top, and a position graph that is the familiar arch.
Prediction
One ball is dropped from rest. At the same instant an identical ball is thrown straight down at 5 metres per second from the same height. Ignore air resistance.
Predict first
How do their accelerations compare while they are falling?
Correct: They are identical
Why: Both are in free fall, so both have exactly the same acceleration of 9.8 metres per second squared downward. The thrown ball is faster the whole way down and lands first, but being faster is a statement about velocity, not acceleration. On a velocity-time graph their lines are parallel: same slope, different starting height, which is precisely the picture that makes this obvious.
Trade off
Both methods are always available. Fill the missing cells to see when each one is less work.
Comparison matrix
| Situation | Graph method | Equation method |
|---|---|---|
| constant velocity, want displacement | one rectangle | also easy, but slower to set up |
| the graph is given and has no numbers | works fine | unusable, nothing to substitute |
| acceleration changes partway through | split into pieces and add areas | must be applied piece by piece too |
| want a precise answer to three digits | hard to read off a sketch | exact |
The graph is better for understanding and for questions with no numbers. The equations are better for precision. Doing a problem both ways, as this deck keeps doing, is the strongest check available to you.
Connect it up
Make the reference sheet you will actually use during homework.
Draw it
On one page, draw three empty axes stacked vertically and label them position, velocity and acceleration. Between each pair, write an arrow pointing down labelled take the slope, and an arrow pointing up labelled take the area. Then, beside each graph, note what a flat line and a steep line mean on that specific graph, because they mean different things on each. Finally add the sign rule: same signs means speeding up, opposite signs means slowing down.
Bring it to the next session. Anything you cannot fill in is exactly what we should spend that hour on.
Exit ticket
The reading that this entire deck was built to make automatic.
Predict first
A velocity-time graph is a straight line sloping downward, still above the time axis for the whole interval shown. What is the object doing?
Correct: Moving forward and slowing down
Why: Above the axis means the velocity is positive, so the object is moving forward the whole time. The downward slope means the velocity is decreasing, so it is slowing down. Reading the height and the slope as two separate pieces of information is the skill; the height gives direction and the slope gives whether it is gaining or losing speed.
Recap
Six things, and two sentences underneath all of them.
| the question asks for | what to do to the graph |
|---|---|
| velocity, from a position graph | take the slope |
| acceleration, from a velocity graph | take the slope |
| displacement, from a velocity graph | take the area, signed |
| distance, from a velocity graph | take the area, sizes only |
| the shape of a position graph | accumulate the area so far |
Slopes go down the ladder and areas go back up it. Every graph question in this unit is one of those two moves, and knowing which one is being asked for is most of the work.
OpenStax, College Physics 2e, Ch. 2 (Kinematics) Ch. 2 — A free second explanation if a topic here needs another pass.
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