1.3 Motion graphs

Little picture stories

Press Next (or Play) to walk through each story one small step at a time. Watch the dot: it draws the graph while the person moves.

1. A walk, and its graph drawn as you go

2. Fast walker, slow walker

On a position-time graph, a flat line means the person is:

3. A mental map for signs

  1. Pick right as positive, before anything else.
  2. Velocity's sign = which way it moves: right +, left −.
  3. Speeding up: acceleration has the same sign as velocity. Slowing down: the opposite sign.
  4. On a velocity–time graph, a rising line = positive acceleration, a falling line = negative, even when the line is below zero.
  5. Distance ignores direction: it only adds up and never goes down.

Two mirrored balls, side by side: Ball A rolls up a ramp to the right, Ball B rolls up a ramp to the left.

The same two balls on a velocity–time graph:

a) Right is positive. A cart moves to the left and is slowing down. What are the signs of its velocity and acceleration?

Show answer

Velocity − (moving left). Slowing down means the opposite sign, so acceleration +.

b) A velocity–time line is below zero and rising toward zero. Is the acceleration positive or negative? Is the object speeding up or slowing down?

Show answer

Rising line: acceleration positive, even below zero. Velocity is negative. Opposite signs, so it is slowing down (its speed shrinks toward zero).

c) You walk 5 m right, then 5 m left. What distance did you walk? Can distance ever go down?

Show answer

5 m + 5 m = 10 m. No: distance only adds up, it never goes down. (Your displacement is 0.)

4. Speeding up draws a curve

5. The trap: 1 m/s² does not mean 1 m in the first second

A ball starts at rest on a gentle slope with an acceleration of 1 m/s2. Many people think it rolls 1 m in the first second. Press Next to step 0.1 s at a time (or Play).

It starts each second slower than it ends it: 0.5 m in second 1, 1.5 m in second 2, 2.5 m in second 3. That is why the x-t graph of speeding up is a curve that gets steeper (card 4).

The bent ramp, and its a-t graph

Now the gentle slope (1 m/s2) is only 3 m long, then the ramp gets steeper (2 m/s2). The ball reaches the bend at ½ × 1 × t2 = 3, so t = √6 ≈ 2.45 s, not 3 s. Watch the acceleration-time graph step up right there.

Check: a cart starts at rest with a = 2 m/s2. How far does it go in the first second? In the first 2 s?

Show answer

First second: ½ × 2 × 12 = 1 m, not 2 m. First 2 s: ½ × 2 × 22 = 4 m. On its x-t graph that is a curve through (1 s, 1 m) and (2 s, 4 m).

An x-t graph is a curve that gets steeper and steeper. The object is:

6. Check yourself

Think of your answer first, then tap to see it.

a) On a position-time graph, the line is flat for a while. What is the person doing?

Show answer

Standing still. Time goes on, but the position does not change.

b) On a position-time graph (position = how far from the door), the line goes down. What is happening?

Show answer

The person is coming back toward the door: moving in the negative direction. A steeper downward line means coming back faster.

c) On a velocity-time graph, the line is flat at +2 m/s. What is the acceleration?

Show answer

0. The velocity does not change, so a = 0. The object moves at a steady 2 m/s in the positive direction.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

On an x-t graph, the slope of the line tells you the:

On a v-t graph, the area under the line tells you the:

A v-t line is below zero and rising. The acceleration is:

Three graphs tell the same story: position, velocity and acceleration against time. Learn to read slope and area and you can turn any one into the other two.

The skateboarder's run

Read the story as text

A coach clips a motion sensor to the end of a long, flat path. A skateboarder pushes off, rolls steadily for a while, drags a foot to slow down, stops, and then rolls back a little toward the sensor.

The sensor does not see a skateboarder. It only prints three lines on a screen.

How do you read the story back out of those lines? And given a story, how do you draw the lines?

The idea

In pictures

  • In the stories, a dot drew the graph while the person moved. Time always runs along the bottom.
  • Position-time: up = moving away, flat = standing still, down = coming back. Steeper = faster.
  • A curve getting steeper = speeding up. A curve levelling off = slowing down.
  • Velocity-time: rising line = positive acceleration, falling = negative, even below zero.
  • The area under a velocity-time graph is how far it went.

Below is the same idea as a table, and then with numbers.

A motion graph is not a picture of the path. Time runs along the bottom. The line shows how one quantity changes as time goes by.

GraphHeight tells youSlope tells youArea under it tells you
x-twhere it isvelocity(nothing useful)
v-tvelocity (sign = direction)accelerationdisplacement Δx
a-tacceleration(not used in AP 1)change in velocity Δv

Slope goes down the list (x → v → a). Area goes up the list (a → v → x). Area below the time axis counts as negative.

Five motions to know by heart

Motionx-tv-ta-t
At restflat lineon zeroon zero
Constant velocity (+)straight line, sloping upflat line above zeroon zero
Speeding up (+ direction)curve getting steeper (bends up)straight line moving away from zeroflat, positive
Slowing down (+ direction)curve levelling off (bends down)straight line heading toward zeroflat, negative
Turning arounda hill (peak) or valleystraight line crossing zeroflat, not zero

The turnaround on an x-t graph is where it is flat for an instant (the top of the hill). On a v-t graph the turnaround is where the line crosses the time axis, not where it has a peak.

Sketching from a story

  1. State which way is positive.
  2. Split the story into parts where the acceleration is constant.
  3. Draw v-t first: for each part, is v positive or negative? Is the speed growing or shrinking? Straight lines only (constant a).
  4. Get a-t from the slope of each v-t piece, and x-t from the area: v positive means x goes up; bigger |v| means steeper.
036 v (m/s) 379t (s) slope = a area = Δx = 39 m

Same idea with numbers

Positive = away from the sensor. The skateboarder's v-t graph above: 0 to 6 m/s in 3 s, steady 6 m/s until 7 s, then down to 0 at 9 s.

Show all steps as text
  1. Acceleration in each part: 6 ÷ 3 = +2 m/s2, then 0, then (0 − 6) ÷ 2 = −3 m/s2. Slope of v-t = rise over run = Δv ÷ Δt.
  2. Displacement, part 1 (triangle): ½ × 3 s × 6 m/s = 9 m. Area of v-t = displacement. A triangle is half base times height.
  3. Part 2 (rectangle): 4 s × 6 m/s = 24 m. Part 3 (triangle): ½ × 2 s × 6 m/s = 6 m.
  4. Total Δx = 9 + 24 + 6 = 39 m. Average velocity = 39 m ÷ 9 s = 4.3 m/s. Check part 1 with kinematics: ½(2)(3)2 = 9 m. Same answer.
  5. x-t shape: curving up (steeper) for 0 to 3 s, straight line for 3 to 7 s, levelling off for 7 to 9 s, then flat. x-t slope = v: it grows, stays the same, then shrinks to zero.

Try it: one motion, three graphs

Pick a preset, predict the position-time graph, then press Play. Watch how the slope of x-t matches the height of v-t, and the slope of v-t matches the height of a-t. Then change the numbers yourself.

position xvelocity vacceleration a

Try this: on "Turning around", step frame by frame to the moment v = 0. The x-t graph is at its peak and flat. The a-t graph is still −2.5 m/s2, not zero.

Practice more: match the graph (move a cart to trace a target graph), slope and area lab, sketching from a story, and the motion graphs practice (ramps, hills and valleys).

Worked examples

Velocity from an x-t line basic

An x-t graph is a straight line from (0 s, 2 m) to (4 s, 10 m). What is the velocity? What do the v-t and a-t graphs look like?

Show solution
Show all steps as text
  1. Slope = (10 − 2) m ÷ (4 − 0) s = +2 m/s. x-t slope = velocity.
  2. A straight line has one slope, so v-t is a flat line at +2 m/s, and a-t is flat on zero.

Area above and below the axis medium

A v-t graph: v = +4 m/s from 0 to 5 s, then a straight line down to −2 m/s at 8 s. Find the acceleration from 5 to 8 s, when the object turns around, its displacement from 0 to 8 s and the distance travelled.

Show solution
Show all steps as text
  1. a = (−2 − 4) ÷ (8 − 5) = −2 m/s2.
  2. Turnaround where v = 0: 4 − 2(t − 5) = 0 → t = 7 s. The line crosses the time axis there.
  3. Areas: 0 to 5 s: 4 × 5 = 20 m. 5 to 7 s: ½(2)(4) = 4 m. 7 to 8 s: ½(1)(−2) = −1 m.
  4. Displacement = 20 + 4 − 1 = +23 m. Distance = 20 + 4 + 1 = 25 m. Area below the axis is negative displacement but still adds to distance.

From a-t to v-t medium

A cart starts from rest. Its a-t graph: +2 m/s2 from 0 to 3 s, 0 from 3 to 5 s, −3 m/s2 from 5 to 7 s. Find v at 3 s, 5 s and 7 s and describe the v-t graph.

Show solution
Show all steps as text
  1. Δv = area of a-t. 0 to 3 s: 2 × 3 = +6 m/s, so v(3) = 6 m/s.
  2. 3 to 5 s: area 0, so v(5) = 6 m/s.
  3. 5 to 7 s: −3 × 2 = −6 m/s, so v(7) = 0.
  4. v-t: straight up from 0 to 6, flat at 6, straight down to 0. The cart stops at 7 s.

Ball rolled up a ramp AP

A ball is given a push up a straight ramp, rolls up, stops, and rolls back down past where it started. Up the ramp is positive. Describe the x-t, v-t and a-t graphs, and say where each graph shows the turnaround.

Show solution
Show all steps as text
  1. The only acceleration (after the push) is down the ramp and constant: a-t is a flat line below zero the whole time, even at the top.
  2. v-t: starts positive and drops in a straight line, crossing zero at the turnaround and continuing below. Constant negative slope = constant negative a.
  3. x-t: an upside-down parabola (a hill). It rises and levels off, is flat at the peak (turnaround), then falls ever more steeply and drops below the start.
  4. Turnaround: top of the x-t hill, the zero crossing of v-t, and nothing special on a-t.

Practice

  1. The slope of a position-time graph gives:

    Show answer

    Slope = Δx ÷ Δt = velocity.

  2. The area between a velocity-time graph and the time axis gives:

    Show answer

    Area = v × Δt = Δx. It gives the change in position; you need the start position to get the final position.

  3. An x-t graph has a positive slope that is getting smaller (the curve bends down and levels off). The object is:

    Show answer

    Positive slope means v > 0. The slope shrinks, so the speed shrinks. v positive and a negative: slowing down.

  4. A v-t graph is a straight line from +8 m/s at t = 0 to 0 at t = 4 s. What is the displacement?

    Show answer

    Triangle area = ½ × 4 s × 8 m/s = 16 m. (−2 m/s2 is the slope, the acceleration; 32 m forgets the ½.)

  5. On a velocity-time graph, how can you tell the object turns around?

    Show answer

    Turning around means the direction of motion flips, so v changes sign. A peak on v-t is just the fastest moment (a = 0 there).

  6. A v-t graph is a straight line sloping upward that starts below zero and ends above zero. Which a-t graph matches?

    Show answer

    A straight v-t line has a constant positive slope, so a is constant and positive the whole time, even while v is negative and even at v = 0.

  7. Short answer (translation between representations). A cart is pushed up a ramp. It slows down, stops, and rolls back down, speeding up. Up the ramp is positive. Describe in words the shapes of the x-t, v-t and a-t graphs from just after the push until it returns to the start, and explain how the three graphs agree at the turnaround.

    Show answer
    • x-t: rises, levels off, is flat at the top, then falls back to the start: a hill shape (concave down).
    • v-t: a straight line from positive down through zero to negative.
    • a-t: a flat line below zero.
    • Agreement: at the top, the x-t slope is zero, which matches v = 0 on v-t. The v-t slope is the same negative number before and after, matching the constant a on a-t. The v-t areas above and below the axis are equal, matching x-t returning to the start.

    Scoring idea: 1 point each for x-t, v-t, a-t shapes, 1 point for a correct slope / area link at the turnaround.

  8. Short answer (mathematical routine). A v-t graph rises in a straight line from 0 to 10 m/s between t = 0 and 5 s, then stays at 10 m/s until 8 s. Find the acceleration in each part, the displacement from 0 to 8 s, and describe the x-t graph.

    Show answer
    1. a = 10 ÷ 5 = 2 m/s2 for 0 to 5 s, and 0 for 5 to 8 s.
    2. Δx = ½(5)(10) + (3)(10) = 25 + 30 = 55 m.
    3. x-t: a curve bending upward (getting steeper) from 0 to 5 s, then a straight line with slope 10 m/s. No corner at 5 s: the slope is 10 m/s on both sides.

AP question types for this topic

Four short free response questions, one of each AP Physics 1 free response type, all about this topic. Write your answer first, then open the worked answer and score yourself with the points shown.

1. Mathematical Routines

Ravi's electric scooter speeds up steadily from rest to a top speed v in time t1, cruises at v for time t2, then brakes steadily to rest in time t3. Its velocity–time graph is a trapezoid.

vt
v–t

(a) Derive an expression for the total displacement in terms of v, t1, t2 and t3.

(b) Calculate it, and the average velocity, for v = 6.0 m/s, t1 = 4.0 s, t2 = 10 s, t3 = 3.0 s.

(c) Find the acceleration in each of the three stages.

Show worked answer and scoring
  1. (a) Displacement = area under the v–t graph = two triangles and a rectangle: Δx = ½v t1 + v t2 + ½v t3. 1 point for using area, 1 point for the expression.
  2. (b) Δx = 12 + 60 + 9 = 81 m. Average velocity = 81 ÷ 17 ≈ 4.8 m/s. 1 point for each.
  3. (c) Slopes: 6.0/4.0 = +1.5 m/s2, then 0, then −6.0/3.0 = −2.0 m/s2. 1 point for all three with signs.

Total: 5 points.

2. Translation Between Representations

Zara's position–time graph, starting from rest, is a curve that starts flat and bends steadily upward (half of a U shape).

vt
A
vt
B
vt
C
vt
D

Velocity–time graphs. A: a straight line rising from zero. B: a flat line. C: a curve bending upward. D: a straight line falling.

(a) Which graph shows Zara's velocity? Justify using the slope of her position graph.

(b) Explain why graph C, which looks just like her position graph, is wrong.

(c) Describe in words her acceleration–time graph.

Show worked answer and scoring
  1. (a) A. Velocity is the slope of the position graph. The slope starts at zero (flat) and grows steadily as the curve bends up, so velocity is a straight line rising from zero. 1 point for A, 1 point for slope reasoning.
  2. (b) C copies the shape of the position graph instead of its slope. A v–t curve bending upward would mean the acceleration keeps growing, which a steady U-shaped curve does not show. 1 point.
  3. (c) A flat line above the axis: constant positive acceleration, equal to the slope of graph A. 1 point for flat, 1 point for positive.

Total: 5 points.

3. Experimental Design and Analysis

Lena gives a cart a push along a level carpet and lets it coast to a stop. A motion sensor at the start records its velocity.

(a) Describe how she should collect data to test whether the carpet slows the cart at a constant rate.

(b) Her readings after the push are below. What graph should she plot, how does it test the idea, and what is the acceleration?

t (s)00.51.01.52.0
v (m/s)1.201.010.790.610.40

(c) Predict when the cart stops and how far it coasts from t = 0.

Show worked answer and scoring
  1. (a) Point the motion sensor along the track, start recording, push the cart briefly by hand away from the sensor, then let go. Use only readings taken after the hand leaves the cart. Repeat a few runs. 1 point for the sensor set-up, 1 point for using only the coasting part.
  2. (b) Plot v against t. A constant slowing rate gives a straight line; these points fall on one, so the idea holds. Slope = (0.40 − 1.20) ÷ 2.0 = −0.40 m/s2. 1 point for v against t, 1 point for the slope.
  3. (c) The line reaches v = 0 at t = 1.20 ÷ 0.40 = 3.0 s. Distance = triangle area = ½ × 1.20 × 3.0 = 1.8 m. 1 point for each.

Total: 6 points.

4. Qualitative/Quantitative Translation

Omar and Aria race. Aria runs at a steady 3.0 m/s from the start line. Omar starts from rest at the same line at the same time, speeding up at 1.2 m/s2. Their position–time graphs cross at t = 5.0 s. Omar claims: “Where our graphs cross, we had the same velocity.”

(a) Explain in words what the crossing point really tells you, and what tells you the velocity.

(b) Write each position as an equation, show they meet at 5.0 s, and find each velocity there.

(c) At what time do they have equal velocities, and how far apart are they then? Connect this to the graphs.

Show worked answer and scoring
  1. (a) Crossing means the same position at the same time. Velocity is the slope, and the two graphs can cross with different steepness. The claim is wrong. 1 point.
  2. (b) xA = 3.0t, xO = 0.60t2. Equal when 3.0t = 0.60t2, so t = 5.0 s (both at 15 m). Then vA = 3.0 m/s but vO = 1.2 × 5.0 = 6.0 m/s. 1 point for the equations, 1 point for the two velocities.
  3. (c) 1.2t = 3.0 gives t = 2.5 s. Gap = 3.0(2.5) − 0.60(2.5)2 = 7.5 − 3.75 = 3.75 m, Aria ahead. On the graphs this is where the two slopes match, which is where the vertical gap between the curves is largest, not where they cross. 1 point for 2.5 s and the gap, 1 point for the link to slopes.

Total: 5 points.

Common mistakes

The mistake: Treating the graph as a picture of the path ("the line goes up, so the cart rolls uphill").

Why it is wrong: The horizontal axis is time, not distance. A cart rolling on flat ground can have any shape of x-t graph.

How to spot it: Read the axis labels first, every time.

The mistake: Reading the height of an x-t graph as the speed.

Why it is wrong: Height on x-t is where the object is. Its speed is the steepness (slope).

How to spot it: A high flat x-t line means "far away and not moving".

The mistake: "v = 0 on the graph, so a = 0."

Why it is wrong: On a v-t line crossing zero, the slope (a) is not zero at the crossing.

How to spot it: Look at the slope of v-t, not its height.

The mistake: Counting area below the time axis as positive displacement.

Why it is wrong: Negative v means moving in the negative direction, so that area takes away from Δx.

How to spot it: Add areas with signs for displacement; add their sizes for distance.

The mistake: Drawing a corner on x-t where the acceleration changes.

Why it is wrong: Velocity cannot jump, so the x-t slope must match on both sides. Only v-t gets corners (when a changes).

How to spot it: At each joint, check that the x-t slope before equals the slope after.

The mistake: "1 m/s2 means it goes 1 m in the first second" (and drawing the x-t graph as a straight line through (1 s, 1 m)).

Why it is wrong: 1 m/s2 means the velocity grows by 1 m/s each second. From rest it goes 0.5 m in the first second, 2.0 m after 2 s and 4.5 m after 3 s: an x-t curve that gets steeper. A 3 m gentle slope is finished at √6 ≈ 2.45 s, not 3 s, and that is where an a-t graph steps up.

How to spot it: From rest use x = ½at2. Check with the v-t graph: the area of the triangle under it is ½ × t × v.