1.2 Displacement, velocity and acceleration

Little picture stories

Press Next (or Play) to walk through each story one small step at a time. Watch the speedometer tags: each one shows what the speedometer read at that spot.

1. Green light: the speedometer climbs

2. Braking is acceleration too

A bike's speedometer reads 8 m/s, then 6 m/s, then 4 m/s, one second apart. What is its acceleration?

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

A ball sits still at the top of a gentle slope. The slope gives it an acceleration of 1 m/s2: its speedometer goes up by 1 m/s every second. How far does it roll in the first second? Many people say 1 m. Press Next to step 0.1 s at a time (or Play) and watch the two counters.

Why only 0.5 m? The ball starts the first second at 0 m/s and only ends it at 1 m/s. On average it went 0.5 m/s, so in 1 s it covers 0.5 m. Each new second starts faster than the last one, so the ball covers 0.5 m, then 1.5 m, then 2.5 m. In short: speed = a × t and total distance = ½ × a × t2.

A bent ramp

Now the gentle slope (1 m/s2) is only 3 m long. After that the ramp bends and gets steeper (2 m/s2). When does the ball reach the bend? Not at 3 s! Step through and watch for the 3 m mark.

½ × 1 × t2 = 3 m gives t2 = 6, so t = √6 ≈ 2.45 s. From that moment the acceleration jumps up from 1 to 2 m/s2 (see the a-t graph in 1.3).

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 (it averages 1 m/s while going from 0 to 2 m/s). First 2 s: ½ × 2 × 22 = 4 m, not 2 × 1 m = 2 m: the second second alone adds 3 m, because the cart starts it already going 2 m/s.

Right is positive. A ball rolls to the left and is speeding up. What is the sign of its acceleration?

5. Check yourself

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

a) A bike rides at 6 m/s. The rider brakes, and the bike loses 3 m/s every second. How fast is it going after 1 s?

Show answer

6 m/s − 3 m/s = 3 m/s. Its acceleration is −3 m/s2: minus, because it points against the motion.

b) Same bike, after 2 s. What is its velocity? What is its acceleration at that moment?

Show answer

Velocity = 3 m/s − 3 m/s = 0: it has just stopped. But the acceleration is still −3 m/s2: right up to that moment the brakes were still taking away 3 m/s every second. Velocity is how fast you go now. Acceleration is how fast that is changing. Zero velocity does not mean zero acceleration. (Once the bike just sits there, nothing changes any more, and then a = 0 too. A ball thrown straight up is the clean case: at the top v = 0, but a = −9.8 m/s2 the whole time.)

c) A car drives at a steady 12 m/s for a whole minute. The speedometer never moves. What is its acceleration?

Show answer

0. The reading never changes, so the velocity changes by 0 m/s each second. Fast does not mean accelerating; changing does.

Already know this?

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

A speedometer reads 0, 3, 6, 9 m/s, one second apart. The acceleration is:

Right is positive. An object has v = −4 m/s and a = +2 m/s². It is:

An object starts at rest with a = 4 m/s². How far does it go in the first second?

How fast, how far, and how quickly the "how fast" is changing. Four equations do almost all of the work.

Green light, then a deer

Part 1: the green light

Part 2: the deer

Read the story as text

You are stopped at a red light. It turns green and you press the gas. Each second your speedometer reads 3 m/s more than the second before.

A few minutes later, on a different road (a dark one), you are cruising at 20 m/s when you see a deer standing on the road 50 m ahead. You brake hard.

How fast are you going 5 s after the light turns green? How far does it take you to stop? Do you hit the deer?

The idea

In pictures

  • Velocity is what the speedometer says right now, plus which way you are going. In the stories it is the green arrow.
  • Acceleration is how much the speedometer reading changes every second. In the stories it is the orange arrow.
  • Reading goes 0, 2, 4, 6: it grows by 2 each second, so the acceleration is 2 m/s each second, written 2 m/s2.
  • Reading goes 10, 8, 6: it shrinks by 2 each second. That is still an acceleration: −2 m/s2, pointing against the motion.
  • Orange arrow the same way as the green arrow: speeding up. Opposite ways: slowing down.

Below is the same idea in words, and then with numbers.

Velocity: how fast position changes

Average velocity is displacement divided by time:

v̄ = Δx ÷ Δt

Instantaneous velocity is the velocity at one moment: what a speedometer would show, plus a sign for direction. You get it by taking Δt very, very short. On a position-time graph it is the slope at that instant (see 1.3).

Acceleration: how fast velocity changes

ā = Δv ÷ Δt     (unit: m/s per s = m/s2)

An acceleration of +3 m/s2 means: every second, the velocity goes up by 3 m/s. Acceleration is a vector, so its sign is a direction too.

Speeding up or slowing down?

Do not look at the sign of a alone. Compare the signs of v and a:

a positivea negativea = 0
v positivespeeding upslowing downsteady speed
v negativeslowing downspeeding upsteady speed

Same signs: speeding up. Opposite signs: slowing down. Picture the arrows: if the acceleration arrow points the same way the object moves, it pushes the speed up.

The kinematic equations (constant acceleration only)

When a is constant, these four equations link x0, x, v0, v, a and t. The first three are on the AP equation sheet.

v = v0 + a t    (no x)

x = x0 + v0 t + ½ a t2    (no v)

v2 = v02 + 2a(x − x0)    (no t)

x − x0 = ½(v0 + v) t    (no a)

How to choose: list what you know and what you want. Pick the equation that is missing the one quantity you do not know and do not need.

Same idea with numbers

Positive = the direction the car drives.

Show all steps as text
  1. At the light: v0 = 0, a = +3.0 m/s2, t = 5.0 s. Want v and Δx. Write the knowns first, with signs.
  2. v = 0 + (3.0 m/s2)(5.0 s) = +15 m/s. v = v0 + a t: the velocity grows 3 m/s every second for 5 s.
  3. Δx = 0 × 5.0 + ½(3.0)(5.0)2 = 37.5 m. x = x0 + v0t + ½at2. Check: average velocity (0 + 15)/2 = 7.5 m/s, times 5 s = 37.5 m.
  4. Braking: v0 = +20 m/s, v = 0, a = −5.0 m/s2. Want the stopping distance. a is negative because it points backward. v and a have opposite signs, so the car slows.
  5. 0 = 202 + 2(−5.0)Δx, so Δx = 400 ÷ 10 = 40 m. No time is given or wanted, so use v2 = v02 + 2aΔx.
  6. Time to stop: t = (0 − 20) ÷ (−5.0) = 4.0 s. The deer is 50 m away, so you stop 10 m short. v = v0 + a t solved for t. Check: (20 + 0)/2 × 4.0 = 40 m.

Try it

position xvelocity vacceleration a

1. Leaving the red light

Predict the velocity-time graph first, then press Play. Change the start velocity and acceleration and predict again.

2. Braking for the deer

The car brakes with a constant (negative) acceleration until it stops. The car does not drive backward: once it stops, it stays stopped. Can you find a speed and braking that hit the deer at 50 m?

Check the equation: the stopping distance is v02 ÷ (2|a|). Double the start speed and the stopping distance becomes four times as long. Try 10 m/s and 20 m/s with a = −5 m/s2: 10 m and 40 m.

More practice: the cart lab lets you set a push and watch v and a change.

Worked examples

A bike speeds up basic

A cyclist goes from 2.0 m/s to 8.0 m/s in 3.0 s along a straight road. What is her average acceleration?

Show solution
Show all steps as text
  1. Positive = direction of travel. Δv = 8.0 − 2.0 = +6.0 m/s.
  2. a = Δv ÷ Δt = 6.0 m/s ÷ 3.0 s = +2.0 m/s2. Same sign as v, so she speeds up. Good.

How long a runway? medium

A plane starts from rest and accelerates at 4.0 m/s2. It needs 80 m/s to take off. How long must the runway be, and how long does the take-off roll last?

Show solution
Show all steps as text
  1. Knowns: v0 = 0, v = 80 m/s, a = 4.0 m/s2. Want Δx (no t given) → use v2 = v02 + 2aΔx.
  2. 6400 = 0 + 2(4.0)Δx, so Δx = 6400 ÷ 8.0 = 800 m.
  3. Time: t = (v − v0) ÷ a = 80 ÷ 4.0 = 20 s. Check: average velocity 40 m/s × 20 s = 800 m.

Slowing down, then turning around medium

A cart on a ramp has v0 = −6.0 m/s and a constant a = +2.0 m/s2. Describe its motion from t = 0 to t = 6.0 s. Find v at 3.0 s and 5.0 s, and the displacement over the 6.0 s.

Show solution
Show all steps as text
  1. At first v is negative and a is positive: opposite signs, so the cart slows down while moving in the negative direction.
  2. v(3.0 s) = −6.0 + 2.0(3.0) = 0. It stops for an instant and turns around. At the turnaround v = 0 but a is still +2.0 m/s2.
  3. v(5.0 s) = −6.0 + 2.0(5.0) = +4.0 m/s. Now v and a are both positive: it speeds up in the positive direction.
  4. Δx(6.0 s) = (−6.0)(6.0) + ½(2.0)(6.0)2 = −36 + 36 = 0. It is back where it started. Constant a makes the motion symmetric about the turnaround: 3 s out, 3 s back.

Reaction time and a stop sign AP

A driver going 25 m/s sees a stop line 60 m ahead. Her reaction time is 0.80 s (the car keeps going at 25 m/s), then she brakes at −6.25 m/s2. Does she stop before the line? If not, how fast is she going at the line?

Show solution
Show all steps as text
  1. Reaction part: constant velocity, Δx1 = 25 × 0.80 = 20 m. No acceleration during the reaction time.
  2. Braking distance to stop: Δx2 = v02 ÷ (2|a|) = 625 ÷ 12.5 = 50 m.
  3. Total = 20 + 50 = 70 m > 60 m. She does not stop in time.
  4. Braking room before the line: 60 − 20 = 40 m. v2 = 625 + 2(−6.25)(40) = 625 − 500 = 125, so v = 11 m/s (11.2 m/s). Use v2 = v02 + 2aΔx for the braking part only.

Practice

  1. An object has v = +4 m/s and a = −2 m/s2. At this moment it is:

    Show answer

    v and a have opposite signs, so it slows down. It is still moving in the + direction because v is positive.

  2. An object has v = −5 m/s and a = −3 m/s2. At this moment it is:

    Show answer

    Same signs mean speeding up. After 1 s, v = −8 m/s: the speed went from 5 to 8 m/s.

  3. A car starts from rest and accelerates at 2.5 m/s2 for 4.0 s. How far does it go?

    Show answer

    Δx = v0t + ½at2 = 0 + ½(2.5)(4.0)2 = 20 m. (40 m forgets the ½; 10 m uses t instead of t2.)

  4. A car moving at 30 m/s brakes at −6.0 m/s2. What is its stopping distance?

    Show answer

    0 = 302 + 2(−6.0)Δx → Δx = 900 ÷ 12 = 75 m. (5.0 s is the stopping time.)

  5. A driver doubles her speed. With the same braking acceleration, her stopping distance becomes:

    Show answer

    Δx = v02 ÷ (2|a|). Distance goes as speed squared: (2)2 = 4.

  6. A ball is thrown straight up. Up is positive. At the very top of its path:

    Show answer

    The velocity is zero for an instant, but it is still changing (from + to −), so the acceleration is not zero. Gravity gives a = −9.8 m/s2 the whole flight. More in 1.5.

  7. Short answer (mathematical routine). A cyclist moving at 12 m/s brakes steadily to 4.0 m/s over a distance of 16 m. Find (a) her acceleration and (b) the time it takes. Check your answer a second way.

    Show answer
    1. Positive = direction of travel. v2 = v02 + 2aΔx: 16 = 144 + 2a(16), so a = (16 − 144) ÷ 32 = −4.0 m/s2.
    2. t = (v − v0) ÷ a = (4.0 − 12) ÷ (−4.0) = 2.0 s.
    3. Check: Δx = ½(v0 + v)t = ½(12 + 4.0)(2.0) = 16 m. It matches.

    Scoring idea: 1 point correct equation with signs, 1 point a with sign and unit, 1 point t, 1 point a valid check.

  8. Short answer (translate words and numbers). A cart has v0 = +8.0 m/s and a = −2.0 m/s2 for 6.0 s. In words, describe what the cart does. Then find its position change and the distance it travels.

    Show answer
    1. It slows down while moving forward (opposite signs), stops at t = 8.0 ÷ 2.0 = 4.0 s, then speeds up backward for 2.0 s.
    2. Δx(6.0 s) = 8.0(6.0) + ½(−2.0)(36) = 48 − 36 = +12 m.
    3. Forward leg (0 to 4 s): ½(8.0 + 0)(4.0) = 16 m. Backward leg (4 to 6 s): ½(2.0)(2.0)2 = 4 m. Distance = 16 + 4 = 20 m. Check: 16 − 4 = 12 m displacement.

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

Elena drives at speed v0 when she sees a fallen branch. She brakes with a constant deceleration of size a until the car stops.

(a) Derive expressions for the stopping time t and the stopping distance d in terms of v0 and a.

(b) Calculate both for v0 = 20 m/s and a = 5.0 m/s2.

(c) With the same brakes she now drives at 40 m/s. Use your expression to find the new stopping distance and explain why it is not just doubled.

Show worked answer and scoring
  1. (a) Final speed 0: 0 = v0 − a t, so t = v0/a. 1 point.
  2. (a) 0 = v02 − 2a d, so d = v02/(2a). 1 point for the right equation, 1 point for solving it.
  3. (b) t = 20/5.0 = 4.0 s; d = 202/(2 × 5.0) = 400/10 = 40 m. 1 point for both with units.
  4. (c) d = 402/10 = 160 m, four times as far. Doubling v0 doubles the time to stop and doubles the average speed while stopping, so the distance grows by 2 × 2 = 4 (it goes as v02). 1 point for 160 m, 1 point for the squared reasoning.

Total: 6 points.

2. Translation Between Representations

Sam gives a cart a push up a smooth ramp. It leaves his hand at +1.5 m/s (up the ramp is positive), slows down, stops for an instant and rolls back down. Its acceleration is a steady −0.50 m/s2 the whole time.

vt
A
vt
B
vt
C
vt
D

Velocity–time graphs. A: a straight line from positive, down through zero, to negative. B: a V shape, down to zero then back up. C: down to zero, then flat on zero. D: a flat line above zero.

(a) Which graph shows the cart's velocity? Justify your choice using the slope.

(b) When does the cart stop, and how far up the ramp does it get?

(c) Describe in words the shape of the acceleration–time graph.

Show worked answer and scoring
  1. (a) A. The slope of a v–t graph is the acceleration, which is constant and negative, so the graph is one straight line sloping down, passing through zero as the cart turns around. B shows speed, not velocity; C has the cart stuck at the top. 1 point for A, 1 point for the slope argument.
  2. (b) t = 1.5 ÷ 0.50 = 3.0 s. Distance = area under the line = ½ × 1.5 × 3.0 = 2.25 m (same as 1.52 ÷ (2 × 0.50)). 1 point for each.
  3. (c) A flat horizontal line below the axis at −0.50 m/s2, the same before, at and after the turn-around. 1 point for flat, 1 point for negative and unbroken at the top.

Total: 6 points.

3. Experimental Design and Analysis

Kofi wants to measure the acceleration of a cart rolling down a slightly tilted track. He has the track, the cart, a metre stick and a stopwatch (or a phone video).

(a) Describe a procedure, saying what he measures and how he makes the start fair.

(b) Starting from rest, d = ½a t2. What should he graph to get a straight line, and what is the slope?

d (m)0.200.400.600.801.00
t (s)0.901.261.561.782.01

(c) Use the data to find the acceleration.

Show worked answer and scoring
  1. (a) Hold the cart at rest at a start line and let go without pushing. Time how long it takes to reach marks 0.20, 0.40, … 1.00 m down the track. Repeat each distance three times and average. 1 point for measuring d and t, 1 point for release from rest and repeats.
  2. (b) Graph d against t2: a straight line through the origin with slope = a/2. 1 point.
  3. (c) t2 = 0.81, 1.59, 2.43, 3.17, 4.04 s2. Best-fit slope ≈ (1.00 − 0.20) ÷ (4.04 − 0.81) = 0.80 ÷ 3.23 ≈ 0.25 m/s2, so a = 2 × 0.25 ≈ 0.50 m/s2. 1 point for t² values, 1 point for a = 2 × slope.

Total: 5 points.

4. Qualitative/Quantitative Translation

Ines claims: “A car that starts from rest with constant acceleration covers three times as far in its second second as in its first second.”

(a) Explain in words, without equations, why the car goes farther in the second second.

(b) Use x = ½a t2 to show that the ratio is exactly 3, whatever a is.

(c) For a = 2.0 m/s2, find the distance in each of the first three seconds and connect the pattern to (a).

Show worked answer and scoring
  1. (a) The car is faster all through the second second than during the first, so its average speed is bigger and it covers more ground in the same time. 1 point.
  2. (b) First second: ½a(1)2 = ½a. Second second: ½a(2)2 − ½a(1)2 = &frac32;a. Ratio = (&frac32;a)/(½a) = 3; a cancels. 1 point for each distance, 1 point for the ratio.
  3. (c) 1st second: 1.0 m; 2nd: 4.0 − 1.0 = 3.0 m; 3rd: 9.0 − 4.0 = 5.0 m. The pattern 1 : 3 : 5 grows by the same step each second because the speed grows by the same amount each second, which is the reason given in (a). 1 point for the numbers, 1 point for the link.

Total: 6 points.

Common mistakes

The mistake: "Negative acceleration means slowing down."

Why it is wrong: A falling ball has a = −9.8 m/s2 (up positive) and it speeds up, because v is negative too.

How to spot it: Compare signs of v and a. Same sign: speeding up. Opposite: slowing down.

The mistake: "When v = 0, a = 0."

Why it is wrong: At a turnaround the velocity is zero for an instant but still changing, so a is not zero.

How to spot it: Ask: is it about to move again? If yes, it is accelerating.

The mistake: Using the kinematic equations when the acceleration changes (for example, across the moment the brakes go on).

Why it is wrong: They are only true for constant a.

How to spot it: Split the motion into parts with one a each (like the reaction time example). The end of one part is the start of the next.

The mistake: Forgetting the ½ or the square in ½at2, or putting a negative a into v2 = v02 + 2aΔx with the wrong sign.

Why it is wrong: The numbers come out 2 or 4 times off, which is exactly what wrong MC choices are built from.

How to spot it: Check with the average velocity: Δx = ½(v0 + v)t.

The mistake: "1 m/s2 means it goes 1 m in the first second."

Why it is wrong: 1 m/s2 means the speed grows by 1 m/s each second. Starting from rest, the speed goes from 0 to 1 m/s during that first second, so the average is 0.5 m/s and it goes only 0.5 m. Then 1.5 m more, then 2.5 m more (2.0 m after 2 s, 4.5 m after 3 s). A 3 m slope is finished at √6 ≈ 2.45 s, not 3 s.

How to spot it: Use x = ½at2 from rest, never x = a × t. Check: average speed × time.