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GCSE Physics · GCSEChapter 16 · Distance, speed, velocity and acceleration
GCSE · Chapter 16

Distance, speed, velocity and acceleration

By the end of this chapter you will be able to

  • Calculate speed, distance and time, and average speed
  • Read a distance-time graph and find the speed from its gradient
  • Find the acceleration from a velocity-time graph, and the distance from the area under it
  • Explain how speed, road conditions and the driver change the stopping distance
  • [H] Use the equations of motion to solve problems with uniform acceleration

You should already know

  • Units of speed and time (KS3 Science and Maths)
  • Forces and Newton’s second law (Chapter 13)

Where this leads

  • Work done and energy (Chapter 15)
  • Momentum (Chapter 14, [H])

GCSE Physics. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Calculate speed as distance divided by time, and distinguish speed from velocity, Interpret distance-time graphs: the gradient is the speed, Interpret velocity-time graphs: the gradient is the acceleration and the area is the distance, Calculate acceleration using a = (v - u) / t, Describe the stopping distance as thinking distance plus braking distance, and the factors that change them, Describe the required practical: the motion of a trolley on a ramp (light gates or ticker timer), [H] Use the equations of motion for uniform acceleration (v = u + at, s = ut + 1/2 a t squared, v squared = u squared + 2as).

Exercise P  Do you remember?

1
FWhat is 100 divided by 20? Write the number.
2
FHow many seconds are there in 3 minutes? Write the number.
3
FWhat is the gradient of a line that rises 6 units for every 2 units across? Write the number.
4
FWhat is 2 x 2 x 2? Write the number.
5
FWhat is the area of a rectangle 10 units wide and 5 units high? Write the number.
16.1

Distance, speed and velocity

GCSE
Calculate speed as distance divided by time, and distinguish speed from velocity
Video lesson: Distance, speed and velocityComing soon
Context. Jo runs round the park. She checks her watch and the distance on her phone. “What is my speed?” she asks. Speed is how fast she moves; velocity also includes the direction she is moving in.

Speed is the distance travelled each second, and is a scalar quantity. It is found from:

speed (m/s) = distance (m) / time (s), so distance = speed x time and time = distance / speed

Velocity is speed in a stated direction. It is a vector. The average speed for a journey is the total distance divided by the total time, not the average of the speeds.

Key ideas

  • Speed = distance / time. Units: m/s (or km/h).
  • Distance = speed x time; time = distance / speed.
  • Velocity = speed in a given direction (a vector).
  • Average speed = total distance / total time.
Worked example 16.1.1
A runner covers 100 m in 20 s. Calculate the speed. Write the number of m/s.

Method. Use speed = distance / time.

  1. Take the numbers. Distance = 100 m and time = 20 s.
    d = 100, t = 20
  2. Divide. Speed = 100 / 20 = 5 m/s.
    100 / 20 = 5
  3. Write the answer.
    5 m/s

Check. 5 x 20 = 100 m. ✓

Worked example 16.1.2
A cyclist rides at 6 m/s for 90 s. How far does she go? Write the number of metres.

Method. Use distance = speed x time.

  1. Take the numbers. Speed = 6 m/s and time = 90 s.
    v = 6, t = 90
  2. Multiply. Distance = 6 x 90 = 540 m.
    6 x 90 = 540
  3. Write the answer.
    540 m

Check. 540 / 90 = 6 m/s. ✓

Worked example 16.1.3
A journey of 120 km takes 1.5 hours. Calculate the average speed in km/h. Write the number of km/h.

Method. Use average speed = total distance / total time.

  1. Read. Distance = 120 km and time = 1.5 h.
    120 km, 1.5 h
  2. Divide. Average speed = 120 / 1.5 = 80 km/h.
    120 / 1.5 = 80
  3. Write the answer.
    80 km/h

Check. 80 x 1.5 = 120 km. ✓

Watch out

Watch out: average speed is total distance over total time. Do not average two speeds if the times are different.

Exam tip

Exam tip. Convert minutes to seconds (or hours) before you calculate, and write the unit with the answer.

Exercise 16A  Speed, distance and time

1
FWhat is the unit of speed commonly used in physics? Write the unit.
2
FA car travels 300 m in 15 s. Calculate its speed. Write the number of m/s.
3
FA train moves at 25 m/s for 40 s. How far does it travel? Write the number of metres.
4
PSA snail moves 12 m in 4 minutes. Calculate its speed in m/s. Write the number.
5
PSA runner runs 2 km in 10 minutes. Calculate the average speed in m/s. Write the number.
6
FA car travels 90 km in 1.5 hours. What is its average speed in km/h? Write the number.
7
FA car drives for 2 hours at 40 km/h. How far does it go? Write the number of km.
8
PSA walker covers 3 km in 45 minutes. What is the average speed in km/h? Write the number.
9
PSA boat goes 200 m north in 50 s, then 200 m south in 50 s. What is its average speed for the whole trip? Write the number of m/s.
10
PSA boat goes 200 m north in 50 s, then 200 m south in 50 s. What is its average velocity for the whole trip? Write the number of m/s.
11
RExplain the difference between speed and velocity, in one sentence. Accept any sentence that says velocity is speed in a stated direction, and speed has no direction.
12
RA pupil says “the average speed is the average of the two speeds in a journey.” Explain why this is wrong, in one sentence. Accept any answer that says it depends on the times, so it should be total distance divided by total time.
13
PSA cyclist travels at 5 m/s for 60 s, then at 10 m/s for 30 s. Find the total distance. Write the number of metres.
16.2

Distance-time graphs

GCSE
Interpret distance-time graphs: the gradient is the speed
Video lesson: Distance-time graphsComing soon
Context. Leo walks to school, stops at the shop, and then runs the rest of the way. His mum plots his distance from home against time. The graph shows each part of the journey, and the steepness of each part shows his speed.

On a distance-time graph, distance is on the vertical axis and time on the horizontal axis. The gradient (the steepness) of a straight line is the speed:

gradient = change in distance / change in time = speed

A horizontal line means the object is stationary (not moving). A steeper line means a higher speed. A curve means the speed is changing.

Distance against time (example)05101520253035400246810time (s)distance (m)
A distance-time graph: a steady walk, a stop, then a faster run

Key ideas

  • Gradient of a distance-time graph = speed.
  • Horizontal line = stationary (speed 0).
  • Steeper line = higher speed.
  • Curved line = changing speed.
Worked example 16.2.1
On the graph, the walk goes from 0 m at 0 s to 20 m at 4 s. Calculate the speed. Write the number of m/s.

Method. Gradient = change in distance / change in time.

  1. Take the numbers. Change in distance = 20 m and change in time = 4 s.
    20, 4
  2. Divide. Speed = 20 / 4 = 5 m/s.
    20 / 4 = 5
  3. Write the answer.
    5 m/s

Check. ✓

Worked example 16.2.2
On the graph, the flat part is from 4 s to 6 s. What is the speed in that time? Write the number of m/s.

Method. A flat line has zero gradient.

  1. Look. The distance stays at 20 m, so the distance does not change.
    20 m
  2. Decide. Gradient = 0 / 2 = 0 m/s, so the object is stationary.
    0
  3. Write the answer.
    0 m/s

Check. ✓

Worked example 16.2.3
On the graph, the run goes from 20 m at 6 s to 40 m at 10 s. Find the speed. Write the number of m/s.

Method. Gradient = change in distance / change in time.

  1. Change. Distance changes by 40 - 20 = 20 m, and time by 10 - 6 = 4 s.
    20, 4
  2. Divide. Speed = 20 / 4 = 5 m/s.
    20 / 4 = 5
  3. Write the answer.
    5 m/s

Check. ✓

Watch out

Watch out: on a distance-time graph, a flat line means stopped, not zero speed in an unusual sense. A steeper line means a faster speed.

Exam tip

Exam tip. To find the gradient, use two points well apart on the line, and read them accurately from the grid.

Exercise 16B  Reading distance-time graphs

1
FWhat does a horizontal line on a distance-time graph show? Write the word: stationary or moving.
2
FOn a distance-time graph, a steeper line means: faster or slower? Write the word.
3
FA line goes from 0 m at 0 s to 30 m at 10 s. What is the speed? Write the number of m/s.
4
PSA line goes from 10 m at 2 s to 50 m at 6 s. What is the speed? Write the number of m/s.
5
PSAn object is at 15 m at 3 s and still at 15 m at 8 s. What is its speed in that time? Write the number of m/s.
6
ROn a distance-time graph, what is the speed at a point where the line is curved, if it is getting steeper? Write the word: increasing or decreasing.
7
PSA line goes from 0 m at 0 s to 60 m at 20 s, then from 60 m at 20 s to 60 m at 30 s. Find the total distance. Write the number of metres.
8
PSUsing the same graph, find the average speed for the whole 30 s. Write the number of m/s.
9
RExplain in one sentence what the gradient of a distance-time line tells you. Accept any sentence that says it gives the speed.
10
PSA line on a distance-time graph goes from 5 m at 1 s to 25 m at 5 s. What is the speed? Write the number of m/s.
16.3

Velocity-time graphs and acceleration

GCSE
Interpret velocity-time graphs: the gradient is the acceleration and the area is the distance, Calculate acceleration using a = (v - u) / t
Video lesson: Velocity-time graphs and accelerationComing soon
Context. A cyclist starts from rest, speeds up to 10 m/s, rides at a steady speed for a while, and then brakes to a stop. A velocity-time graph shows all three stages in one picture.

On a velocity-time graph, velocity is on the vertical axis and time on the horizontal axis.

Acceleration is in metres per second squared (m/s2). A negative acceleration is a deceleration (slowing down).

Velocity against time (example)02468100246810time (s)velocity (m/s)
A velocity-time graph: speeding up, constant velocity, then slowing down to a stop

Key ideas

  • a = (v - u) / t: u = start velocity, v = final velocity, t = time.
  • Gradient of a velocity-time graph = acceleration.
  • Area under a velocity-time graph = distance.
  • Constant velocity: horizontal line, zero acceleration.
Worked example 16.3.1
A car speeds up from 2 m/s to 10 m/s in 4 s. Calculate the acceleration. Write the number of m/s squared.

Method. Use a = (v - u) / t.

  1. Take the numbers. u = 2 m/s, v = 10 m/s, t = 4 s.
    u = 2, v = 10, t = 4
  2. Calculate. a = (10 - 2) / 4 = 8 / 4 = 2 m/s2.
    8 / 4 = 2
  3. Write the answer.
    2 m/s2

Check. ✓

Worked example 16.3.2
On the velocity-time graph, the cyclist moves at 10 m/s for 4 s (from 4 s to 8 s). Find the distance travelled in this part. Write the number of metres.

Method. The area of a rectangle is velocity x time.

  1. Area. Distance = 10 x 4 = 40 m.
    10 x 4 = 40
  2. Check. A horizontal line gives a rectangle under the graph.
    rectangle
  3. Write the answer.
    40 m

Check. ✓

Worked example 16.3.3
Find the distance covered in the first 4 s, when the velocity rises from 0 to 10 m/s in a straight line. Write the number of metres.

Method. The area under the sloping line is a triangle.

  1. Base and height. Base = 4 s, height = 10 m/s.
    4, 10
  2. Area. Area = 0.5 x 4 x 10 = 20 m.
    0.5 x 4 x 10 = 20
  3. Write the answer.
    20 m

Check. ✓

Watch out

Watch out: acceleration is the gradient, not the height of the graph. A high velocity with a flat line has zero acceleration.

Exam tip

Exam tip. For distance on a velocity-time graph, split the area into rectangles and triangles, add them up, and check the units.

Exercise 16C  Acceleration and velocity-time graphs

1
FWhat is the unit of acceleration? Write the unit.
2
FA car accelerates from 0 to 12 m/s in 6 s. Calculate its acceleration. Write the number of m/s squared.
3
PSA ball slows down from 8 m/s to 2 m/s in 3 s. Calculate its acceleration. Write the number, with a minus sign if it is a deceleration.
4
PSA train accelerates at 0.5 m/s squared from rest for 20 s. What is its final velocity? Write the number of m/s.
5
FA velocity-time graph has a horizontal line at 15 m/s for 8 s. How far is travelled in that time? Write the number of metres.
6
PSA cyclist speeds up uniformly from 0 to 6 m/s in 3 s. How far does she travel in those 3 s? Write the number of metres.
7
FA velocity-time graph rises from 0 to 20 m/s in 10 s. What is the gradient? Write the number of m/s squared.
8
FOn a velocity-time graph, the line is horizontal at 0 m/s. What does this show? Write the word: rest or motion.
9
PSA car at 15 m/s brakes at 3 m/s squared. How long does it take to stop? Write the number of seconds.
10
RExplain in one sentence why the area under a velocity-time graph gives the distance. Accept any sentence that says velocity times time gives distance, which is the area.
11
PSA cyclist decelerates from 12 m/s to rest in 4 s, at a constant rate. Find the distance travelled while braking. Write the number of metres.
12
PSA velocity-time graph shows a constant speed of 4 m/s from 0 s to 10 s, then 4 m/s to 0 m/s in 2 s. Find the total distance. Write the number of metres.
16.4

Stopping distances

GCSE
Describe the stopping distance as thinking distance plus braking distance, and the factors that change them
Video lesson: Stopping distancesComing soon
Context. A driver sees a child run into the road. Her foot takes a moment to move to the brake, and then the car takes time to slow down. The total distance the car travels is the stopping distance.

The stopping distance is the thinking distance plus the braking distance:

stopping distance = thinking distance + braking distance

The thinking distance is the distance travelled in the driver’s reaction time. It depends on the speed and on the driver. The braking distance is the distance travelled while the brakes slow the car. It depends on the speed, the brakes, the tyres and the road.

speed (m/s)thinking distance (m)braking distance (m)stopping distance (m)
106.05.011.0
159.011.220.2
2012.020.032.0
2515.031.246.2
3018.045.063.0
Example stopping distances: thinking time 0.6 s and braking deceleration 10 m/s squared
FactorWhat it does to stopping distance
speedthinking distance rises in proportion; braking distance rises much faster (about speed squared)
tiredness or alcohol or drugslonger thinking time, so a longer thinking distance
road conditions (wet, icy)less grip, so a longer braking distance
brakes or tyres in poor conditionless braking force, so a longer braking distance
Factors that change the stopping distance

Key ideas

  • Stopping distance = thinking + braking.
  • Thinking distance rises with speed and with tiredness, alcohol or drugs.
  • Braking distance rises much faster with speed (about the speed squared).
  • Road conditions and worn brakes or tyres increase the braking distance.
Worked example 16.4.1
Using the example table, a car at 20 m/s has a thinking distance of 12 m and a braking distance of 20 m. What is the stopping distance? Write the number of metres.

Method. Add the two distances.

  1. Read. Thinking distance = 12 m and braking distance = 20 m.
    12, 20
  2. Add. Stopping distance = 12 + 20 = 32 m.
    12 + 20 = 32
  3. Write the answer.
    32 m

Check. ✓

Worked example 16.4.2
Using the example table, how many times bigger is the braking distance at 30 m/s than at 10 m/s? Write the number.

Method. Divide the two braking distances.

  1. Read. At 10 m/s the braking distance is 5 m; at 30 m/s it is 45 m.
    5 and 45
  2. Divide. 45 / 5 = 9 times.
    45 / 5 = 9
  3. Write the answer.
    9

Check. 9 = 3 squared, so the braking distance rises with speed squared. ✓

Worked example 16.4.3
Using the example table, how many metres further does a car at 30 m/s travel to stop than one at 20 m/s? Write the number of metres.

Method. Find both stopping distances, then subtract.

  1. Read. At 30 m/s the stopping distance is 63 m; at 20 m/s it is 32 m.
    63 and 32
  2. Subtract. Difference = 63 - 32 = 31 m.
    63 - 32 = 31
  3. Write the answer.
    31 m

Check. ✓

Watch out

Watch out: the braking distance rises much faster than the speed. Tripling the speed makes the braking distance nine times bigger, not three times.

Exam tip

Exam tip. When explaining a factor, say whether it changes the thinking distance, the braking distance, or both, and why.

Exercise 16D  Stopping distances

1
FWhat is the name of the distance travelled during the driver’s reaction time? Write the phrase.
2
FWhat is the name of the distance travelled while the brakes are slowing the car? Write the phrase.
3
FStopping distance = thinking distance + ... Write the missing word.
4
FName one factor that increases the thinking distance. Write the word.
5
FName one factor that increases the braking distance. Write the word.
6
FUsing the example table, what is the thinking distance at 15 m/s? Write the number of metres.
7
PSUsing the example table, what is the stopping distance at 25 m/s? Write the number of metres to 1 decimal place.
8
RExplain why the braking distance increases faster than the speed, in one sentence. Accept any sentence that says the kinetic energy depends on speed squared, so more energy must be removed by the brakes.
9
RA driver is tired. Explain how this affects the stopping distance, in one sentence. Accept any sentence that says the thinking distance increases because the reaction time is longer.
10
PSAt 10 m/s, a thinking distance of 6 m and a braking distance of 5 m. At 20 m/s, the braking distance is 20 m. How many times bigger is the braking distance at 20 m/s than at 10 m/s? Write the number.
11
PSA driver has a thinking distance of 9 m and braking distance of 11.25 m at 15 m/s. What is the stopping distance? Write the number of metres.
12
RWhich would increase the stopping distance more: a 10% increase in speed, or a 10% increase in thinking time? Write speed or thinking time.
16.5

Required practical: trolley on a ramp

GCSE
Describe the required practical: the motion of a trolley on a ramp (light gates or ticker timer)
Video lesson: Required practical: trolley on a rampComing soon
Context. The class releases a trolley at the top of a ramp and measures how fast it is moving at two points. They want to find the acceleration of the trolley down the ramp.

Method (with a teacher). Set up a ramp at a slight slope, with the trolley held at the top. Attach a card of known width to the trolley, and place two light gates a known distance apart. Release the trolley and record the time for the card to pass each gate. The speed at each gate is card width / time. The acceleration is (v - u) / t, where t is the time between the two gates. (A ticker timer works in the same way, using the spacing of dots on the tape.)

Safety. Keep the trolley away from the edge of the bench, and stop it before it falls off the ramp. Wear shoes and keep feet clear.

trolley on a ramp, with two light gatesgate 1gate 2a flag or card on the trolley breaks each gate
A trolley on a ramp, timed by two light gates

Key ideas

  • Speed at a gate = width of card / time for the card to pass.
  • Acceleration = (v - u) / t between two gates.
  • Use the same card on the trolley every time.
  • Plot speed against time: the gradient gives the acceleration.
Worked example 16.5.1
A card 0.05 m wide takes 0.10 s to pass gate 1, and 0.05 s to pass gate 2. Find the speed at gate 2. Write the number of m/s.

Method. Speed = width / time.

  1. Read. Card width = 0.05 m and time = 0.05 s.
    0.05, 0.05
  2. Calculate. Speed at gate 2 = 0.05 / 0.05 = 1 m/s.
    0.05 / 0.05 = 1
  3. Write the answer.
    1 m/s

Check. ✓

Worked example 16.5.2
Gate 1 gives a speed of 0.5 m/s at 1.0 s, and gate 2 gives 1.5 m/s at 3.0 s. Find the acceleration. Write the number of m/s squared.

Method. Use a = (v - u) / t.

  1. Change. Change in speed = 1.5 - 0.5 = 1.0 m/s.
    1.0
  2. Time. Time between gates = 3.0 - 1.0 = 2.0 s.
    2.0
  3. Calculate. a = 1.0 / 2.0 = 0.5 m/s2.
    1.0 / 2.0 = 0.5
  4. Write the answer.
    0.5 m/s2

Check. ✓

Worked example 16.5.3
A card 0.02 m wide takes 0.04 s to pass a light gate. Find the speed at the gate. Write the number of m/s.

Method. Speed = width / time.

  1. Read. Card width = 0.02 m and time = 0.04 s.
    0.02, 0.04
  2. Calculate. Speed = 0.02 / 0.04 = 0.5 m/s.
    0.02 / 0.04 = 0.5
  3. Write the answer.
    0.5 m/s

Check. ✓

Watch out

Watch out: check the time between the gates, not the total time from the release. Use the same card throughout, because its width is needed for each speed.

Exam tip

Exam tip. Describe the method, the measurements, and how the results are used: speed from width and time, acceleration from the gradient or from (v - u) / t.

Exercise 16E  The trolley on a ramp

1
FIn the ramp practical, how is the speed at a gate found? Write the equation in words.
2
RWhy is the same card used every time? Write one reason in one sentence. Accept any sentence that says the card width is needed for each speed, so it must be the same.
3
PSA card of width 0.04 m passes a gate in 0.08 s. Calculate the speed. Write the number of m/s.
4
PSThe speed at gate 1 is 0.2 m/s at 0.5 s, and at gate 2 it is 0.6 m/s at 1.5 s. Find the acceleration. Write the number of m/s squared.
5
RA ticker timer makes 50 dots per second. The tape shows 10 dots in 0.2 s. Describe how you would find the speed from the tape. Accept any answer that says measure the distance between the dots and divide by the time between them.
6
RIn the ramp practical, why must the ramp be at a slight slope rather than flat? Write one reason in one sentence. Accept any sentence that says gravity needs to give a component of force along the ramp so the trolley accelerates.
7
FA pupil plots speed against time for the trolley and gets a straight line through the origin. What does the gradient show? Write the word: acceleration or distance.
8
FName a suitable device, other than light gates, for timing the trolley on the ramp. Write the name.
9
RExplain why a longer ramp gives more accurate times, in one sentence. Accept any sentence that says the trolley is timed over a longer distance, so the timing error is a smaller fraction.
10
PSIn a practical, the speed at gate 2 is 1.2 m/s and the acceleration is 0.6 m/s squared. The gates are 2 s apart. Find the speed at gate 1. Write the number of m/s.
16.6

[H] Equations of motion for uniform acceleration

GCSE
[H] Use the equations of motion for uniform acceleration (v = u + at, s = ut + 1/2 a t squared, v squared = u squared + 2as)
Video lesson: [H] Equations of motion for uniform accelerationComing soon

[H] Higher tier only. For an object moving with uniform (constant) acceleration, three equations link the variables: u = initial velocity, v = final velocity, a = acceleration, t = time, and s = displacement (distance in a straight line). They are:

v = u + a t, s = ((u + v) / 2) x t, s = u t + ½ a t2, v2 = u2 + 2 a s

Choose the equation that contains the three quantities you know and the one you want.

Key ideas

  • [H] v = u + a t.
  • [H] s = u t + 1/2 a t squared.
  • [H] v squared = u squared + 2 a s.
  • [H] Use the equation with the four quantities that you know or want.
Worked example 16.6.1
[H] A car starts at 2 m/s and accelerates at 2 m/s squared for 4 s. Find its final velocity. Write the number of m/s.

Method. Use v = u + a t.

  1. Take the numbers. u = 2 m/s, a = 2 m/s2, t = 4 s.
    u = 2, a = 2, t = 4
  2. Calculate. v = 2 + 2 x 4 = 10 m/s.
    2 + 8 = 10
  3. Write the answer.
    10 m/s

Check. ✓

Worked example 16.6.2
[H] Using the same car, find the distance travelled in those 4 s. Write the number of metres.

Method. Use s = u t + 1/2 a t squared.

  1. Take the numbers. u = 2, a = 2, t = 4.
    u = 2, a = 2, t = 4
  2. Calculate. s = 2 x 4 + 0.5 x 2 x 4 squared = 8 + 16 = 24 m.
    8 + 16 = 24
  3. Write the answer.
    24 m

Check. ✓

Worked example 16.6.3
[H] A car at 10 m/s decelerates at 2 m/s squared until it stops. Find the distance it travels. Write the number of metres.

Method. Use v squared = u squared + 2 a s.

  1. Take the numbers. u = 10 m/s, v = 0, a = -2 m/s2.
    u = 10, v = 0, a = -2
  2. Rearrange. s = (v squared - u squared) / (2a) = (0 - 100) / (2 x -2) = 25 m.
    25
  3. Write the answer.
    25 m

Check. ✓

Watch out

Watch out: [H] a deceleration is a negative acceleration. Use a = -2, not 2, when the car is slowing down.

Exam tip

Exam tip. [H] Write down u, v, a, t and s, fill in what you know, and then choose the equation with only one unknown.

Exercise 16F  [H] Equations of motion

1
F[H] A car starts from rest and accelerates at 3 m/s squared for 5 s. Find its final velocity. Write the number of m/s.
2
PS[H] A trolley moves at 4 m/s and accelerates at 2 m/s squared for 3 s. Find its final velocity. Write the number of m/s.
3
PS[H] An object starts from rest and accelerates at 4 m/s squared for 3 s. Find the distance travelled. Write the number of metres.
4
PS[H] A car accelerates from 5 m/s to 25 m/s over 100 m. Find its acceleration. Write the number of m/s squared.
5
PS[H] A car at 20 m/s brakes at 4 m/s squared. Find its stopping distance. Write the number of metres.
6
PS[H] A ball is thrown upwards and has an initial speed of 9.8 m/s, with a deceleration of 9.8 m/s squared. How long to reach its highest point? Write the number of seconds.
7
R[H] Explain why the equation s = u t + 1/2 a t squared can be used only for uniform acceleration, in one sentence. Accept any sentence that says the acceleration must be constant for the equations to hold.
8
PS[H] A car starts from rest and accelerates at 2 m/s squared for 6 s. Find the distance travelled. Write the number of metres.
9
PS[H] A trolley at 10 m/s decelerates at 5 m/s squared until it stops. Find its stopping distance. Write the number of metres.
10
PS[H] A cyclist accelerates from 2 m/s to 8 m/s in 3 s. Find the acceleration. Write the number of m/s squared.

Chapter summary

SkillMethod
Speedspeed = distance / time; average speed = total distance / total time
Distance-time graphGradient = speed; horizontal = stationary
Velocity-time graphGradient = acceleration; area = distance
Accelerationa = (v - u) / t
Stopping distanceThinking + braking; both increase with speed, and braking rises fastest
[H] Equations of motionv = u + at; s = ut + 1/2 at squared; v squared = u squared + 2as

Can you…?

  • I can calculate speed, distance and time
  • I can read a distance-time graph and find the speed
  • I can find acceleration from a velocity-time graph and the distance from its area
  • I can explain how factors change the stopping distance
  • I can describe the trolley and ramp practical
  • [H] I can use the equations of motion

Key terms

speed
distance travelled per unit time (a scalar)
velocity
speed in a stated direction (a vector)
acceleration
the rate of change of velocity, in m/s squared
thinking distance
the distance travelled in the driver’s reaction time
braking distance
the distance travelled while the brakes slow the vehicle
[H] uniform acceleration
acceleration that stays the same throughout

Exercise 16Z  Check-up

1
A cyclist covers 360 m in 20 s. What is her speed? Write the number of m/s.
2
A car moves at 12 m/s for 30 s. How far does it go? Write the number of metres.
3
A velocity-time graph has a gradient of 3 m/s squared. What does this mean? Write the number of m/s squared for the acceleration.
4
A car accelerates from 4 m/s to 16 m/s in 6 s. Calculate its acceleration. Write the number of m/s squared.
5
What is the stopping distance if the thinking distance is 6 m and the braking distance is 14 m? Write the number of metres.
6
Name the factor that makes the braking distance longer on wet roads. Write the word.
7
A line on a distance-time graph is horizontal. What is the speed? Write the number of m/s.
8
A ball speeds up from rest to 6 m/s in 3 s. How far does it travel, if the speed rises uniformly? Write the number of metres.
9
Explain why a car travels further while braking if it is going faster, in one sentence. Accept any sentence that says the braking distance depends on speed, and more kinetic energy must be removed.
10
[H] A car starts at 0 and accelerates at 2 m/s squared for 10 s. Find the distance. Write the number of metres.

Total 29 marks. Allow about 29 minutes.

Exercise 16X  Exam-style questions

1
PSA runner covers 800 m in 2 minutes 40 seconds. Calculate the average speed in m/s. Write the number.[3 marks]
2
PSA velocity-time graph rises from 0 to 12 m/s in 4 s, then stays at 12 m/s for 6 s. Calculate the total distance travelled. Write the number of metres.[3 marks]
3
PSA distance-time graph shows a line from 0 m at 0 s to 45 m at 15 s, then a horizontal line until 25 s. Calculate the average speed over the 25 s. Write the number of m/s.[2 marks]
4
RDescribe how a student can use light gates to find the acceleration of a trolley on a ramp. Accept any answer that describes timing the card passing each gate, calculating each speed, then using a = (v - u) / t.[4 marks]
5
RExplain why the braking distance of a car increases much faster than its speed, using the idea of kinetic energy. Accept any answer that says kinetic energy is proportional to speed squared, so the brakes must remove much more energy, which needs a longer distance.[4 marks]
6
PSA car has a thinking distance of 15 m and a braking distance of 30 m at its speed. Calculate the stopping distance. Then state how the stopping distance changes if the driver is tired (longer reaction time). Write the stopping distance in metres.[2 marks]
7
PS[H] A car decelerates uniformly from 30 m/s to rest in 5 s. Calculate the deceleration and the distance travelled while stopping. Write two values: the deceleration in m/s squared (as a positive number), then the distance in metres, separated by a comma.[4 marks]
8
PS[H] A trolley starts from rest on a ramp and accelerates at 0.4 m/s squared for 5 s. Calculate its speed after 5 s, and the distance it travels. Write two values: the speed in m/s, then the distance in metres, separated by a comma.[3 marks]
9
PSA car accelerates uniformly from rest to 20 m/s in 8 s. Calculate the distance travelled in this time. Write the number of metres.[2 marks]
10
RDescribe how a light gate and a card can be used to measure the speed of a trolley at one point. Accept any answer that says the card width divided by the time for the card to pass the gate.[2 marks]
← Chapter 15: Work done, energy and springsChapter 17: Waves: transverse, longitudinal and wave speed →

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