Answers · Acceleration, falling and drag

Acceleration — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. FQ1 · Practice

Model answer. Δv = final speed − starting speed = 8.0 − 2.0 = 6.0 m/s. Acceleration a = Δv ÷ Δt = 6.0 m/s ÷ 4.0 s = 1.5 m/s². Her speed grows by 1.5 m/s in every second.

Checked automatically

Acceleration of the cyclist: 1.5 m/s² (within ±2%)
- Common wrong answer 2.0 : You divided the final speed by the time. Use the change in speed: 8.0 − 2.0 = 6.0 m/s.
- Common wrong answer 24 : You multiplied. Acceleration = change in speed ÷ time.
- Common wrong answer 2.5 : You added the two speeds. The change in speed is final − starting: 8.0 − 2.0 = 6.0 m/s.

A full answer includes

I found the change in speed: Δv = 8.0 − 2.0 = 6.0 m/s.
I divided by the time: a = 6.0 m/s ÷ 4.0 s = 1.5 m/s², with the unit m/s².

Also accept

1.5 m/s²

Watch for

“a = 8.0 ÷ 4.0 = 2.0 m/s².” — 8.0 m/s is her final speed, not the change. She started at 2.0 m/s, so her speed changed by 6.0 m/s.
“a = 6.0 × 4.0 = 24 m/s².” — Acceleration is the change in speed per second, so divide the change by the time.

2. FQ2 · Practice

Model answer. c. Car Y's speed grows by 0.4 m/s in every second (0, 0.4, 0.8, 1.2, 1.6 m/s): a constant acceleration of 0.4 m/s². Car X's speed grows by 1.2, then 0.8, then 0.4, then 0.2 m/s: it is still speeding up, but by less each second, so it has a changing acceleration. Car X is faster, but that does not make its acceleration constant.

A full answer includes

I chose c.
I worked out the change in speed in each second: car Y gains 0.4 m/s every second, so it has a constant acceleration; car X gains 1.2, 0.8, 0.4 and 0.2 m/s, so it has a changing acceleration that gets smaller.

Also accept

c; Y gains the same 0.4 m/s each second, X gains less each second.

Watch for

“Car X has the constant acceleration because it goes faster.” — How fast a car goes says nothing about whether its speed changes steadily. Look at the change in speed in each second.
“Both speeds go up every second, so both accelerations are constant.” — Constant acceleration means the speed goes up by the same amount every second. Car X's gains get smaller.

3. FQ3 · Practice

Model answer. Section A rises from 0 to 6 m/s between 0 s and 3 s. Gradient = rise ÷ run = 6 m/s ÷ 3 s = 2.0 m/s². The gradient of a speed–time graph is the acceleration, so her acceleration is 2.0 m/s².

Checked automatically

Acceleration in section A: 2.0 m/s² (within ±2%)
- Common wrong answer 0.5 : You divided the time by the speed. Gradient = rise (speed) ÷ run (time).
- Common wrong answer 6 : That is the speed at the end of section A, not the acceleration. Divide the rise by the run.
- Common wrong answer 18 : You multiplied. Gradient = rise ÷ run.

A full answer includes

I read the rise, 6 − 0 = 6 m/s, and the run, 3 − 0 = 3 s.
I divided: gradient = 6 m/s ÷ 3 s = 2.0 m/s², because the gradient of a speed–time graph is the acceleration.

Also accept

2.0 m/s² (2 m/s²)

Watch for

“Her acceleration is 6 m/s², because the line reaches 6.” — 6 m/s is a speed. Acceleration is the change in speed divided by the time the change took.
“Gradient = 3 ÷ 6 = 0.5 m/s².” — Rise ÷ run: the rise is up the speed axis (6 m/s), the run is along the time axis (3 s).

4. FQ4 · Practice

Model answer. Δv = final − starting = 5.0 − 15 = −10 m/s. a = Δv ÷ Δt = −10 m/s ÷ 4.0 s = −2.5 m/s². The acceleration is negative, so it is a deceleration: the speed falls by 2.5 m/s in every second. (b) The force is friction between the pads and the discs. It acts against the motion of the discs, so it slows the wheels and the bus down, and it heats the discs and the pads.

Checked automatically

(a) Acceleration of the bus: -2.5 m/s² (within ±2%)
- Common wrong answer 2.5 : Right size, wrong sign. The speed gets smaller, so Δv = 5.0 − 15 is negative, and so is the acceleration.
- Common wrong answer 3.75 : You divided the starting speed by the time. Use the change in speed: final − starting.
- Common wrong answer 5 : You added the speeds. The change in speed is final − starting: 5.0 − 15 = −10 m/s.

A full answer includes

I found Δv = 5.0 − 15 = −10 m/s.
I divided: a = −10 m/s ÷ 4.0 s = −2.5 m/s². The minus sign shows a deceleration of 2.5 m/s²: the bus is slowing down.
(b) Friction: it acts against the turning of the discs, so it slows the wheels and the bus, and it heats the discs.

Also accept

−2.5 m/s² (a deceleration of 2.5 m/s²); (b) friction: slows the discs, heats them

Watch for

“a = +2.5 m/s², because an acceleration is always positive.” — The bus is slowing down, so the change in velocity is negative and so is the acceleration: −2.5 m/s², a deceleration.
“a = −2.5 m/s² means the bus is moving backwards.” — The bus keeps moving forwards. The minus sign says its speed is getting smaller.

5. FQ5 · Practice

Model answer. a = Δv ÷ Δt, so Δv = a × Δt = 9.8 m/s² × 2.0 s = 19.6 m/s. The stone started from rest, so after 2.0 s its speed is 19.6 m/s (about 20 m/s).

Checked automatically

Speed after 2.0 s: 19.6 m/s (within ±2.5%)
- Common wrong answer 9.8 : That is the speed gained in one second. The stone has been falling for 2.0 s.
- Common wrong answer 4.9 : You divided by the time. Speed gained = acceleration × time.

A full answer includes

I used change in speed = acceleration × time = 9.8 m/s² × 2.0 s = 19.6 m/s.
It started from rest, so its speed after 2.0 s is 19.6 m/s (20 m/s to 2 significant figures).

Also accept

19.6 m/s (20 m/s)

Watch for

“Its speed is 9.8 m/s, the value of g.” — g is an acceleration: the speed gained in each second. After 2.0 s the stone has gained twice that.
“A heavier stone would be going faster after 2.0 s.” — In free fall every object has the same acceleration, so every stone has the same speed after 2.0 s.

6. FQ6 · Practice

Model answer. (a) Acceleration is the change in velocity per unit time: a = Δv / Δt. (b) Δv = 80 − 0 = 80 m/s, so a = 80 m/s ÷ 32 s = 2.5 m/s². (c) Its speed grows by 2.5 m/s in each second: 2.5 m/s after 1 s, 5.0 m/s after 2 s, and so on, up to 80 m/s after 32 s.

A full answer includes

(a) Acceleration is the change in velocity per unit time: a = Δv / Δt.
(b) a = (80 − 0) m/s ÷ 32 s = 2.5 m/s², with the unit m/s².
(c) The aircraft's speed grows by 2.5 m/s in every second.

Also accept

(a) change in velocity per unit time; (b) 2.5 m/s²; (c) the speed grows by 2.5 m/s every second

Watch for

“(a) Acceleration is how fast something goes.” — That is speed. Acceleration is how quickly the speed changes.
“(c) It travels 2.5 m in every second.” — 2.5 m/s² is not a distance. It is the speed the aircraft gains in each second.

7. FQ7 · Practice

Options as numbered on the pupil sheet: 1 — Constant acceleration; 2 — Changing acceleration: speeding up by less each second; 3 — Constant speed: zero acceleration; 4 — Constant deceleration.

Model answer. A: a straight rising line, so the speed grows by the same amount every second: constant acceleration. B: a curve that gets flatter, so the motorbike still speeds up, but by less each second: a changing acceleration that gets smaller. C: a flat line at 10 m/s, so the speed does not change: zero acceleration. D: a straight falling line, so the speed falls by the same amount every second: a constant deceleration.

A full answer includes

I matched A: constant acceleration; B: changing acceleration, speeding up by less each second; C: constant speed, zero acceleration; D: constant deceleration.

Also accept

A constant acceleration; B changing acceleration; C constant speed; D constant deceleration.

Watch for

“C: at rest, because the line is flat.” — On a speed–time graph a flat line at 10 m/s means moving at a steady 10 m/s. At rest would be flat along the time axis, at 0.
“B: constant acceleration, because the speed is still going up.” — The speed goes up by less each second, so the acceleration is changing. A constant acceleration gives a straight line.

8. FQ8 · Practice

Model answer. (a) Section A rises from 4 m/s to 16 m/s in 6 s: a = 12 m/s ÷ 6 s = 2.0 m/s². (b) Section B is flat at 16 m/s: constant speed, acceleration 0. (c) Section C falls from 16 m/s to 0 between 12 s and 17 s: a = (0 − 16) m/s ÷ 5 s = −3.2 m/s². The acceleration is −3.2 m/s², so the deceleration is 3.2 m/s².

A full answer includes

(a) Gradient = (16 − 4) m/s ÷ (6 − 0) s = 12 m/s ÷ 6 s = 2.0 m/s².
(b) The line is flat, so the speed does not change: the acceleration is zero.
(c) Gradient = (0 − 16) m/s ÷ (17 − 12) s = −16 m/s ÷ 5 s = −3.2 m/s²: a deceleration of 3.2 m/s².

Also accept

(a) 2.0 m/s²; (b) 0; (c) −3.2 m/s², deceleration 3.2 m/s²

Watch for

“(a) 16 ÷ 6 = 2.7 m/s².” — The car started at 4 m/s, not from rest. Use the change in speed, 16 − 4 = 12 m/s.
“(c) 16 ÷ 17 = 0.94 m/s².” — Use the time the braking took, 17 − 12 = 5 s, and the change in speed, 0 − 16 = −16 m/s.

9. FQ9 · Practice

Model answer. a, b and c. Where the blocks rub on the rims there is friction, a force between solid surfaces that acts against the motion, so the wheels slow down (a). Friction heats the surfaces, so the rims and the blocks get warmer (b). The speed falls, so Δv is negative and so is the acceleration: a deceleration (c). Friction can also stop something moving in the first place, such as a heavy box that does not slide when you push it gently, so d is wrong. A negative acceleration means slowing down, not moving backwards, so e is wrong.

A full answer includes

I chose a, b and c.
I explained that friction between the blocks and the rims acts against the motion and slows the wheels, and that it heats the surfaces; I did not choose d or e.

Also accept

a, b, c

Watch for

“e: a negative acceleration means the bicycle moves backwards.” — The bicycle keeps moving forwards while it slows down. The minus sign means its speed is getting smaller.
“d: friction only acts on things that are moving.” — Push a heavy box gently and it may not slide at all: friction can stop motion starting.

10. FQ10 · Practice

Model answer. b. Near the Earth every object in free fall has the same acceleration, the acceleration of free fall g, about 9.8 m/s². a is wrong: without air a heavy ball and a feather fall together. c is wrong: 9.8 m/s² is an acceleration, so the speed grows by 9.8 m/s in every second instead of staying the same. d is wrong: g is approximately constant, but it is very slightly different in different places, a little smaller at the top of a high mountain.

A full answer includes

I chose b.
I explained that a is wrong because without air every object has the same acceleration; c is wrong because 9.8 m/s² is an acceleration, so the speed grows by 9.8 m/s every second; d is wrong because g is only approximately constant: it is very slightly different in different places.

Also accept

Watch for

“a: heavier objects fall faster.” — Without air every object falls with the same acceleration. A feather falls slowly in air because the air pushes back on it.
“c: in free fall an object falls at a steady 9.8 m/s.” — The unit is m/s², an acceleration: the speed grows by 9.8 m/s in every second.

11. FT1 · Apply

Model answer. (a) Taking upwards as positive, the velocity falls by 9.8 m/s every second, so a = −9.8 m/s²: a deceleration of 9.8 m/s². (b) The ball stops rising when its velocity reaches 0: Δv = 0 − 14.7 = −14.7 m/s, and Δt = Δv ÷ a = −14.7 ÷ −9.8 = 1.5 s. (c) A straight line from 14.7 m/s at 0 s down to 0 at 1.5 s: a constant deceleration gives a straight line with a negative gradient. (d) No. At the top its velocity is zero for an instant, but its velocity is still changing at 9.8 m/s every second: a moment earlier it was moving up, a moment later it moves down. Its acceleration is still 9.8 m/s² downwards, the acceleration of free fall. (e) Friction between the match head and the rough side of the box. It acts against the motion, so the match only keeps moving while you push it; and friction heats the rubbing surfaces, here enough for the head to catch fire.

A full answer includes

(a) −9.8 m/s²: its speed gets smaller by 9.8 m/s every second, a deceleration.
(b) Δv = 0 − 14.7 = −14.7 m/s, so Δt = Δv ÷ a = −14.7 m/s ÷ −9.8 m/s² = 1.5 s.
(c) Labelled axes (time / s, speed / m/s) and a straight line falling from 14.7 m/s at 0 s to 0 at 1.5 s.
(d) No. Its velocity is zero for an instant, but it is still changing: just before the top it moves up, just after it moves down. Gravity still acts, so the acceleration is still 9.8 m/s² downwards.
(e) Friction between two solid surfaces: it acts against the motion of the match, so you must keep pushing, and it heats the surfaces until the head is hot enough to catch fire.

Also accept

(a) −9.8 m/s²; (b) 1.5 s; (c) straight line from 14.7 m/s to 0 at 1.5 s; (d) no, still 9.8 m/s² downwards; (e) friction: acts against the motion, heats the head

Watch for

“(d) At the top the acceleration is zero because the ball has stopped.” — Zero velocity is not zero acceleration. The velocity is still changing: up just before, down just after.
“(b) 14.7 × 9.8 = 144 s.” — The speed falls by 9.8 m/s each second. How many seconds does it take to lose 14.7 m/s?