Full chapter
Kinematics
All 6 topics and the revision summary on one page.
01
Distance, speed and velocity
Make a guess. You are not marked.
A runner covers the first 100 m of a track at 5 m/s and the next 100 m at 10 m/s. Is the average speed for the 200 m equal to 7.5 m/s?
Show the answer
No. It is less than 7.5 m/s.
The slow half takes 20 s and the fast half only 10 s, so the runner spends twice as long at 5 m/s. Average speed = 200 m / 30 s = 6.7 m/s.
Distance is the total length of the path travelled. Displacement is the change in position from start to finish, including direction.
If you walk 3 m to the right, then 1 m back to the left, you have walked 4 m in total, but finish 2 m to the right of where you started. Your distance is 4 m; your displacement is 2 m to the right.
- Speed
- Distance travelled per unit time. It tells you how fast, without a direction. For example, 2 m/s.
- Velocity
- Rate of change of displacement. It tells you how fast and in which direction. For example, 2 m/s to the right.
Speed is a scalar; velocity is a vector. A scalar has magnitude (size) only, while a vector has magnitude and direction. Along a straight line, choose one direction as positive. Motion in the opposite direction has negative velocity, even though speed is always non-negative.
The time includes any stops. Average speed describes the whole journey; the object does not have to travel at that speed at every instant. Convert units before substituting: 1 minute = 60 s and 1 km = 1000 m. To convert km/h to m/s, divide by 3.6.
Distance in centimetres divided by time in seconds gives cm/s. For example, 30 cm in 2.0 s gives 15 cm/s = 0.15 m/s, because each centimetre is 0.01 m.
Worked example
Average speed including a stop
A student walks 60 m in 40 s, waits for 20 s, then walks another 40 m in 40 s. Find the average speed.
- Add the distances: 60 + 40 = 100 m.
- Include all the time: 40 + 20 + 40 = 100 s.
- Divide: average speed = 100 m / 100 s = 1 m/s.
This means that travelling steadily at 1 m/s for 100 s would cover the same total distance.
Try it step by step
Work it out in steps: Total distance in m, then Total time in s, then Average speed in m/s.
Show the answer
Problem 1.
Total distance in m: Distance = 300 + 180 = 480 m
Total time in s: Time = 50 + 10 + 40 = 100 s (the stop counts)
Average speed in m/s: Average speed = 480 / 100 = 4.8 m/s (2 sf)
Problem 2.
Total distance in m: Distance = 120 + 80 = 200 m
Total time in s: Time = 15 + 25 + 20 = 60 s
Average speed in m/s: Average speed = 200 / 60 = 3.3 m/s (2 sf)
Problem 3.
Total distance in m: Distance = 900 + 600 = 1500 m
Total time in s: Time = 60 + 30 + 60 = 150 s
Average speed in m/s: Average speed = 1500 / 150 = 10 m/s (2 sf)
Include the whole journey. Add all the distances and all the times, including stops. Taking a simple mean of the speeds only works when the time spent at each speed is equal.
Check your understanding
A cyclist travels 90 m in 15 s, waits for 5 s, then travels 30 m in 10 s. What is the average speed for the whole journey?
Show the answer
4 m/s
Total distance is 120 m and total time is 30 s, so average speed = 120 / 30 = 4 m/s.
02
Acceleration and slowing down
Make a guess. You are not marked.
A car is moving forwards. Its acceleration is -3 m/s2, with forwards taken as positive. What is the car doing?
Show the answer
Still moving forwards, but slowing down.
A negative acceleration means the velocity is decreasing by 3 m/s every second. The car keeps moving forwards until its velocity reaches zero.
Acceleration is the change in velocity per unit time.
Velocity includes direction. If that distinction is unfamiliar, review speed and velocity first.
Uniform acceleration means equal changes in velocity in equal time intervals. If velocity rises from 0 to 2 to 4 to 6 m/s at one-second intervals along a straight line, the acceleration is 2 m/s2. Every second adds another 2 m/s of velocity.
This gives the acceleration throughout an interval if it is uniform. If it varies, the calculation gives the average acceleration over that interval.
Worked example
Acceleration while slowing down
A trolley moving forwards slows uniformly from 8 m/s to 2 m/s in 3 s. Take forwards as positive.
a = (2 - 8) / 3 = -2 m/s2
The velocity decreases by 2 m/s each second. It still moves forwards throughout, because its velocity stays positive. We can also say it has a deceleration of 2 m/s2.
Try it step by step
Work it out in steps: Change in velocity v - u in m/s, then Acceleration (v - u) / t in m/s2.
Show the answer
Problem 1.
Change in velocity v - u in m/s: v - u = 16 - 4 = 12 m/s
Acceleration (v - u) / t in m/s2: a = 12 / 6 = 2 m/s2
Problem 2.
Change in velocity v - u in m/s: v - u = 5 - 25 = -20 m/s
Acceleration (v - u) / t in m/s2: a = -20 / 8 = -2.5 m/s2 (the minus sign means it is slowing)
Problem 3.
Change in velocity v - u in m/s: v - u = 30 - 0 = 30 m/s
Acceleration (v - u) / t in m/s2: a = 30 / 12 = 2.5 m/s2
For non-uniform acceleration, the velocity changes by different amounts in equal times. A car leaving a junction may gain 2 m/s in its first second, then 3 m/s in the next, then only 1 m/s as the driver eases off. It speeds up in each interval, but not at a constant rate.
Negative acceleration does not always mean slowing. If velocity changes from -2 to -5 m/s, speed increases from 2 to 5 m/s. Velocity and acceleration point in the same negative direction. A body slows when acceleration opposes its velocity.
Watch out for this
A student says: "A car with a negative acceleration must be moving backwards." What is wrong with this?
Show the answer
Negative acceleration means the acceleration points in the negative direction. A car moving forwards (positive velocity) with negative acceleration is slowing down. It moves backwards only if its velocity is negative.
Check your understanding
At 0, 1, 2 and 3 s, a car moving forwards has velocities of 0, 2, 5 and 9 m/s. Is its acceleration uniform?
Show the answer
Not uniform: the velocity rises by 2, then 3, then 4 m/s.
The average accelerations in the three one-second intervals are 2, 3 and 4 m/s2. The acceleration is not constant.
03
Displacement-time graphs
Make a guess. You are not marked.
On a graph of distance against time, the line curves upwards, getting steeper as time goes on. What is the object doing?
Show the answer
It is speeding up, because the gradient, its speed, keeps increasing.
The gradient of a distance-time graph is the speed. A line that gets steeper has an increasing gradient, so the object is speeding up.
Read displacement from the height and velocity from the gradient.
Use the definitions of displacement and velocity. A point such as (2 s, 3 m) pairs a time on the horizontal axis with a reading on the vertical axis.
Time goes on the horizontal axis. Displacement from the chosen starting point goes on the vertical axis. The gradient is the slope of the line: the vertical change divided by the horizontal change. Here it tells you how much displacement changes per second.
At rest
Uniform velocity
Non-uniform velocity
Worked example
Calculating a gradient
A straight segment joins (2 s, 3 m) to (5 s, 15 m).
Gradient = (15 - 3) / (5 - 2) = 12 / 3 = 4 m/s
The object moves at a constant velocity of +4 m/s during this interval. Dividing 15 by 5 would use the origin, which is not on this segment.
A curve has a changing gradient. Where it becomes steeper, the object is moving faster. Where it becomes flatter, it is slowing down. To estimate velocity at one instant, draw a tangent that follows the curve's direction at that point, then calculate the tangent's gradient. A line joining two points on the curve gives an average over that interval.
Worked example
Finding the gradient at one instant
Find the velocity at 2 s on this curved graph. Draw a straight tangent that follows the curve's direction at the marked point, (2 s, 1 m).
- Choose two well-separated points on the tangent: A is (1 s, 0 m) and B is (4 s, 3 m).
- Vertical change = 3 - 0 = 3 m. Horizontal change = 4 - 1 = 3 s.
- Gradient = 3 m / 3 s = 1 m/s.
The estimated velocity is +1 m/s at 2 s. The 3 s interval belongs to the gradient triangle; it does not turn this into an average over the curved motion.
A downward-sloping displacement-time line has negative velocity: the object is moving in the chosen negative direction. A negative displacement tells you which side of the origin it is on, not which direction it is moving. Check the gradient for its direction of motion.
A higher line does not necessarily mean faster motion. Compare gradients, not heights. A horizontal line above zero still represents an object at rest.
04
Velocity-time graphs
Height gives velocity. Gradient gives acceleration. Area gives displacement.
Here a gradient measures change in velocity per second. It has a different meaning from the gradient of a displacement-time graph because the vertical quantity has changed.
The vertical axis now shows velocity at each time. A horizontal line on this graph means constant velocity. The object is at rest only if that line is at zero.
| Graph shape | Motion |
|---|---|
| Line at zero | At rest throughout that interval. |
| Horizontal line above zero | Constant positive velocity; zero acceleration. |
| Straight sloping line | Uniform acceleration. Its gradient is constant. |
| Curve | Non-uniform acceleration. Its gradient changes. |
A curve becoming less steep still shows a changing velocity, but the magnitude of acceleration is getting smaller. For a sloping straight segment, calculate acceleration using the change in velocity divided by the elapsed time. In the one-direction journey below, forwards is positive.
A curve means changing acceleration
This example moves forwards throughout. Its graph becomes steeper: the velocity increases by a larger amount each second.
| Time / s | Velocity / (m/s) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
Over the successive one-second intervals, the changes are 1, 3 and 5 m/s. The average accelerations over those intervals are therefore 1, 3 and 5 m/s2. The acceleration is non-uniform. A straight rising line would instead have equal changes in equal times.
Why area gives displacement
At constant velocity, displacement = velocity x time. On this graph, the height is velocity and the width is time, so their product is the area of a rectangle. When velocity changes uniformly, use the area under the sloping line, splitting it into rectangles and triangles where needed.
For displacement, count area above the time axis as positive and area below it as negative. For total distance, add the sizes of all the areas without cancelling them.
Motion and graphs
Moving in one direction
A trolley moves along a straight track in one direction. It travels 18 m in 6 s.
Move the slider to see the trolley's position and the values on both graphs at the selected time. You can also use the arrow keys.
Moving forwards at a constant 4 m/s.
- Displacement
- 6 m
- Velocity
- 4 m/s
- Acceleration
- 0 m/s2
View the data table
| Time / s | Displacement / m | Velocity / (m/s) |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 1 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 4 |
| 4 | 12 | 4 |
| 5 | 16 | 4 |
| 6 | 18 | 0 |
Worked example
Find the distance and average speed
The trolley reaches 4 m/s in 2 s, stays at 4 m/s for 3 s, then brakes to rest in 1 s. Use the one-direction example above. Split the area under the graph into two triangles and a rectangle.
- 0 to 2 s, triangle: (1/2) x 2 s x 4 m/s = 4 m.
- 2 to 5 s, rectangle: 3 s x 4 m/s = 12 m.
- 5 to 6 s, triangle: (1/2) x 1 s x 4 m/s = 2 m.
Total distance = 4 + 12 + 2 = 18 m
Average speed = 18 / 6 = 3 m/s
Multiplying time by speed gives metres: s x m/s = m. All velocities here are positive, so displacement is also +18 m.
Find the accelerations from the gradients
First interval: (4 - 0) / 2 = 2 m/s2.
Middle interval: (4 - 4) / 3 = 0 m/s2.
Final interval: (0 - 4) / 1 = -4 m/s2.
The trolley loses 4 m/s of velocity each second while braking, compared with a gain of 2 m/s each second at the start.
Try it step by step
Work it out in steps: Area of the triangle in m, then Area of the rectangle in m, then Total distance in m.
Show the answer
Problem 1.
Area of the triangle in m: Triangle: 1/2 x 4 x 6 = 12 m
Area of the rectangle in m: Rectangle: 5 x 6 = 30 m
Total distance in m: Distance = 12 + 30 = 42 m
Problem 2.
Area of the triangle in m: Triangle: 1/2 x 5 x 8 = 20 m
Area of the rectangle in m: Rectangle: 10 x 8 = 80 m
Total distance in m: Distance = 20 + 80 = 100 m
Problem 3.
Area of the triangle in m: Triangle: 1/2 x 6 x 12 = 36 m
Area of the rectangle in m: Rectangle: 4 x 12 = 48 m
Total distance in m: Distance = 36 + 48 = 84 m
Pure / Changing direction
Distance and displacement when direction changes
Select Changing direction above to follow this example. Velocity falls uniformly from +2 to -2 m/s over 4 s. The trolley slows to rest at 2 s, turns, then speeds up backwards.
- Acceleration: (-2 - 2) / 4 = -1 m/s2 throughout.
- Signed area before 2 s: (1/2) x 2 x 2 = +2 m.
- Signed area after 2 s: -(1/2) x 2 x 2 = -2 m.
- Displacement: +2 + (-2) = 0 m. The trolley returns to its start.
- Total distance: 2 + 2 = 4 m. Add the magnitudes of both areas.
Average speed is 4 / 4 = 1 m/s; average velocity is 0 / 4 = 0 m/s. At the turning instant, velocity is zero but acceleration is still -1 m/s2. There is no pause lasting a whole interval.
Check the vertical axis. Area under a displacement-time graph does not give displacement. The area rule here works because the vertical axis is velocity.
Check your understanding
On a velocity-time graph, the line is horizontal at +3 m/s for 4 s. What happens?
Show the answer
Acceleration is zero and displacement is +12 m.
The gradient is zero. The rectangular area is 3 m/s x 4 s = 12 m. The object keeps moving throughout.
05
Free fall
Make a guess. You are not marked.
A hammer and a feather are released together from the same height inside a chamber with all the air pumped out. Which reaches the floor first?
Show the answer
They land at the same time.
With no air, gravity is the only force on each object, and both accelerate at the same g regardless of mass. Same start, same acceleration, same landing time.
An object is in free fall when gravity is the only force acting on it.
Use a = (v - u) / t to connect the initial velocity, final velocity and elapsed time. Choose a positive direction before assigning signs.
Near the Earth's surface, the acceleration of free fall is approximately constant at 10 m/s2 downwards. It is represented by g. Use this model when air resistance is negligible (small enough to ignore), and use the value of g given in the question.
A ball released from rest gains about 10 m/s of downward velocity each second. Its velocity is about 10 m/s downwards after 1 s, and 20 m/s downwards after 2 s, provided it has not yet hit the ground.
Worked example
A ball dropped from rest
A ball falls for 0.60 s near Earth. Ignore air resistance and take downwards as positive, with g = 10 m/s2. Find its velocity.
- Initial velocity: u = 0 because the ball is released from rest.
- Rearrange a = (v - u) / t: v = u + at.
- Substitute: v = 0 + (10 x 0.60) = 6.0 m/s downwards.
The ball gains 6.0 m/s of downward velocity in 0.60 s. Its acceleration remains 10 m/s2; g is an acceleration, not a velocity.
Try it step by step
Work it out in steps: Change in velocity g x t in m/s, then Final velocity v = u + gt in m/s.
Show the answer
Problem 1.
Change in velocity g x t in m/s: g x t = 10 x 1.2 = 12 m/s
Final velocity v = u + gt in m/s: v = u + gt = 0 + 12 = 12 m/s downwards
Problem 2.
Change in velocity g x t in m/s: g x t = 10 x 2.0 = 20 m/s
Final velocity v = u + gt in m/s: v = 3.0 + 20 = 23 m/s downwards
Problem 3.
Change in velocity g x t in m/s: g x t = 10 x 0.80 = 8.0 m/s
Final velocity v = u + gt in m/s: v = 0 + 8.0 = 8.0 m/s downwards
Free fall does not mean no force. Gravity still acts. Two objects released together from rest at the same height have the same acceleration in this model, even if their masses differ. Real air resistance can make their motions different.
If air resistance is significant, the acceleration may change as the object falls. Falling with air resistance explains the changing force balance and terminal velocity.
Check your understanding
Two small balls of different masses are dropped from rest at the same height. If air resistance is negligible, which explanation is correct?
Show the answer
Both gain about 10 m/s of downward velocity each second.
Both have approximately the same constant downward acceleration. Released together from rest, they fall together in this model.
Optional: measuring g with light gates (not needed for the exam)
Measure free-fall acceleration with two light gates
A short opaque flag passes through two stationary light gates as a compact body falls. Each gate times how long the same known flag length blocks its beam. This gives two speeds, which can be compared over their matching time interval.
Only the flag crosses the beams
- Measure the flag length parallel to its travel. Align the same flag with both beams and secure the gates. The body and attachment must neither add another interruption nor strike a gate.
- Connect the gates to a compatible timer or logger. Select a mode that records each blocked duration and the times of those intervals on the same clock.
- Release the body above the measured region so that it travels vertically without appreciable rotation. Use the catch below. Record several falls to inspect the spread of the readings.
Use a supplied set of readings
In this example, L = 20.0 mm = 0.0200 m. The length instrument resolves 0.1 mm and the logger resolves 0.0001 s. The clock's zero is the start of the first blockage; the body is already moving then.
| Reading | Gate 1 | Gate 2 |
|---|---|---|
| Block begins / s | 0.0000 | 0.1070 |
| Block ends / s | 0.0200 | 0.1170 |
| Blocked duration / s | 0.0200 | 0.0100 |
| Midpoint time / s | 0.0100 | 0.1120 |
| Average speed during blockage / m/s | 0.0200 / 0.0200 = 1.00 | 0.0200 / 0.0100 = 2.00 |
Length divided by blocked duration gives the average speed during that short interval. The body keeps moving down, so these are also positive downward velocities. For uniform acceleration, each interval's average velocity equals the velocity at its time midpoint.
Match the interval to the two speed readings
Both rows share one linear time scale. Shaded windows show the blocked intervals; the dots mark their midpoints.
Calculate the acceleration
Matching time interval = 0.1120 - 0.0100 = 0.1020 s.
Acceleration = change in velocity / matching time interval
= (2.00 - 1.00) / 0.1020 = 9.80392... m/s2.
Report about 9.8 m/s2 downward, consistent with using 10 m/s2 as the usual approximate value. The gate separation is not needed for this flag-timing calculation.
The 0.1070 s between the two starts, and the 0.0870 s gap between the blockages, describe different intervals. Neither matches these midpoint velocities.
Choose instruments and check the model
- Length: choose an instrument whose range covers the whole flag and whose resolution supports a 20.0 mm reading. For these supplied readings it resolves 0.1 mm. Check its zero and measure along the travel direction.
- Timing: use a compatible electronic mode and range that capture the whole event and resolve 0.0100 s blockages. The example resolution is 0.0001 s. A hand-operated stopwatch cannot usefully time such a short interruption.
- Alignment and forces: keep the flag aligned, avoid rotation or contact, and use a compact body and short flag to limit air resistance. Check that the effective optical interruption length agrees sufficiently with the measured length.
Call the measured acceleration g only while gravity is the only significant force after release. Air resistance or contact can invalidate the free-fall interpretation. In less uniform motion, a short-interval average is only an approximation to a local velocity; extra timer digits do not make it an exact instantaneous value.
Repeated falls reveal variation, but do not correct a wrong flag length, systematic beam misalignment or a wrongly chosen time interval. Improve the cause of the error; timing resolution alone is not accuracy.
Check your understanding
Two short light-gate intervals give downward speeds of 1.00 and 2.00 m/s. Their midpoint times are 0.0100 and 0.1120 s. For approximately uniform acceleration, which calculation uses the correct time interval?
Show the answer
(2.00 - 1.00) / 0.1020, giving about 9.8 m/s2 downwards.
The velocities belong to the interval midpoints, so their matching separation is 0.1120 - 0.0100 = 0.1020 s. Divide the change in velocity by that interval.
06
Plotting motion graphs
Make a guess. You are not marked.
A car accelerates uniformly from rest at 2.0 m/s2 for 5.0 s, then travels at a steady speed for 5.0 s. Which part covers more distance?
Show the answer
The steady part: 50 m, against 25 m while accelerating.
Accelerating: area under the speed-time line = 1/2 x 5.0 x 10 = 25 m. Steady at 10 m/s for 5.0 s: 50 m.
Use the readings to plot a graph, then use its gradients to describe the motion.
This example connects the displacement-time gradient with the velocity-time height and area. Rectangle area is width x height; triangle area is half of that.
A trolley moves forwards at a constant 2 m/s for 2 s, waits for 2 s, then moves forwards at a constant 3 m/s for another 2 s. Its recorded displacements are shown below. We model the changes between stages as abrupt.
| Time / s | Displacement / m |
|---|---|
| 0 | 0 |
| 1 | 2 |
| 2 | 4 |
| 3 | 4 |
| 4 | 4 |
| 5 | 7 |
| 6 | 10 |
Plot the displacement-time graph
- Put time on the horizontal axis and displacement on the vertical axis. Label both quantities and units.
- Choose uniform scales that use the space: 0 to 6 s and 0 to 10 m cover these readings.
- Plot the pairs. Join each stage with a straight line because the motion is stated to be uniform within that stage. For a smoothly changing motion, use an appropriate smooth curve.
- Check the interpretation: rising from 0 to 2 s, flat from 2 to 4 s, then a steeper rise from 4 to 6 s.
Worked example
Sketch the velocity-time graph
Use the gradients you just found: draw horizontal segments at 2 m/s from 0 to 2 s, at 0 from 2 to 4 s, and at 3 m/s from 4 to 6 s. Mark the abrupt changes at the boundaries; a real trolley takes time to change its velocity.
The area is (2 x 2) + (0 x 2) + (3 x 2) = 10 m. This agrees with the final displacement on the first graph. Average speed is 10 / 6 = 1.67 m/s to 3 significant figures.
When interpreting experimental data: a few readings tell you average motion between those times. They do not prove exactly what happened between measurements. Here the description supplies the assumption of constant speed within each stage. Repeated readings and sensible scales help you spot an anomalous value before drawing a trend.
Try another set of readings
This optional task uses a different trolley, moving forwards at a constant speed within each stage. Sketch both motion graphs, describe the three stages and find the total distance. You can use paper; there is no need to submit an answer.
| Time / s | Displacement / m |
|---|---|
| 0 | 0 |
| 2 | 6 |
| 4 | 6 |
| 6 | 8 |
Show the worked answer
On the displacement-time graph, join (0, 0) to (2, 6), then to (4, 6), then to (6, 8) with straight segments. Label the time axis in seconds and the vertical axis in metres.
The gradients are (6 - 0) / 2 = 3 m/s, (6 - 6) / 2 = 0 m/s, and (8 - 6) / 2 = 1 m/s. The trolley moves quickly, stops for 2 s, then moves more slowly.
On the velocity-time graph, draw horizontal segments at 3, 0 and 1 m/s over the same intervals. Its area is (3 x 2) + (0 x 2) + (1 x 2) = 8 m.
During the stop, the first graph stays at 6 m while the second stays at zero. Both lines are horizontal, but their vertical axes describe different quantities.
Revision summary
- Average speed
- Total distance / total elapsed time. Include stops. Unit: m/s.
- Acceleration
- (v - u) / t. Constant throughout a uniform interval; otherwise an average. Unit: m/s2.
- Free fall near Earth
- g is approximately 10 m/s2 downwards. Air resistance is neglected.
| Feature | Displacement-time | Velocity-time |
|---|---|---|
| Height | Displacement in m | Velocity in m/s |
| Gradient | Velocity | Acceleration |
| Horizontal line | At rest | Constant velocity; rest only at zero |
| Area | Does not give distance or displacement | Displacement; add the sizes of all areas for distance |
Measure free-fall acceleration
Two light gates time the passage of the same short flag. Speed = flag length / blocked duration. For approximately uniform acceleration, assign each speed to the midpoint of its blocked interval, then use the time between those midpoints in a = (v - u) / t. The supplied speeds 1.00 and 2.00 m/s at 0.0100 and 0.1120 s give about 9.8 m/s2 downwards. This is a free-fall determination only if other forces are negligible. Align the flag and gates, use a suitable timer and keep the attached body out of the beam; repeats do not fix a wrong flag length or time interval.
Check your answer
- Identify the axes and the time interval.
- Use changes on both axes for a gradient.
- Split graph areas into rectangles and triangles; triangle area = (1/2) x base x height.
- Give units, and direction when the quantity needs it.
- State assumptions such as uniform acceleration or negligible air resistance.
- Separate negative velocity, negative acceleration and slowing down. They do not mean the same thing.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
A cyclist starts from rest and accelerates uniformly to 6.0 m/s in 4.0 s. She then rides at a constant 6.0 m/s for a further 10 s. Calculate the total distance she travels in the 14 s. [2]
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Exam-style question
Define acceleration. A car speeds up uniformly from 12 m/s to 30 m/s in 6.0 s. Calculate its acceleration. [3]
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Exam-style question
A girl walks in a straight line away from her home. A graph of her distance from home against time has three straight parts: from (0 s, 0 m) to (40 s, 60 m), then flat from 40 s to 70 s, then from (70 s, 60 m) to (100 s, 150 m). Describe her motion in each part and calculate her speed in the last part. [3]
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Exam-style question
A stone is dropped from rest from a bridge and hits the water 1.5 s later. Air resistance is negligible and the acceleration of free fall is 10 m/s2. Calculate the speed of the stone as it hits the water and the height of the bridge above the water. [3]
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Quick recall
A cyclist travels 90 m in 15 s, waits for 5 s, then travels 30 m in 10 s. What is the average speed for the whole journey?