Speed, velocity and distance-time graphs, acceleration, Newton's second law, stopping distances and terminal velocity, built up from everyday examples.
Objectives
Calculate speed, distance and average speed, and read a distance-time graph.
Calculate acceleration from a change in velocity and a time.
Read acceleration and constant velocity from a velocity-time graph.
Use resultant force = mass × acceleration (Newton's second law) and describe terminal velocity.
Explain what makes up a stopping distance, what changes it and how reaction time is measured.
Use v² − u² = 2as, tangents and areas under graphs (Higher tier).
Picture a cyclist waiting at a red light. The light turns green, she pushes the pedals and her speed climbs. A few minutes later she sees a dog, squeezes the brakes and stops. Speeding up and slowing down are what this lesson is about.
Think about a rollercoaster. At the top of the first drop it hardly moves, then it shoots down and builds up speed very quickly. Your stomach notices that sudden change. Physics has a word for how quickly speed changes, and that word is acceleration.
Forces are the reason speed changes. A bigger push makes the speed change faster, and a heavier object needs a bigger push to change by the same amount. We will turn that into a short equation and use it on cars and skydivers.
Along the way you will meet stopping distance, which decides how far a car travels before it can stop. You will see why a skydiver reaches a top speed and then stops speeding up. Both ideas are about forces balancing or not balancing.
By the end you will be able to do four things. You will work out speed and acceleration, and read distance-time and velocity-time graphs. You will use force = mass × acceleration. You will explain terminal velocity. You will say what makes a stopping distance and what changes it.
A small note on words. Speed is how fast you go. Velocity is speed in a stated direction. In a straight line the two are almost the same thing, so do not worry if the two words feel close.
Speed, velocity and distance-time graphs
Speed is how fast something is moving. It is the distance travelled every second. Speed has a size and no direction, so it is a scalar. Velocity is speed in a given direction, such as 5 m/s north, so it is a vector.
Speeds change from person to person and from place to place. Age, fitness, the ground and the distance all matter. Typical values are a walk at about 1.5 m/s, a run at about 3 m/s and a cycle at about 6 m/s. A car in town moves at about 13 m/s, which is 30 mph, and on a motorway at about 30 m/s. A passenger plane flies at about 250 m/s. Sound in air travels at about 330 m/s, and the speed of the wind changes from day to day.
For an object moving at a constant speed, distance travelled = speed × time. In symbols, s = v × t. The distance is in metres, the speed is in m/s and the time is in seconds. A walker at 1.5 m/s for 60 s goes 1.5 × 60 = 90 m. Rearrange it to find the speed, v = s ÷ t, or the time, t = s ÷ v.
Speed is rarely constant. When you walk, run or ride in a car, you keep speeding up and slowing down. So we often use the average speed, which is total distance ÷ total time. A runner covers 400 m in 80 s. Her average speed is 400 ÷ 80 = 5 m/s, even though she was faster at some moments and slower at others.
A distance-time graph shows a journey. Time goes along the bottom and distance goes up the side. A flat line means the object is stationary. A straight sloping line means a constant speed, and a steeper line means a faster speed. A curve that gets steeper means the object is speeding up.
The gradient of a distance-time graph is the speed. Gradient = change in distance ÷ change in time. In the graph, the first part rises 30 m in 10 s, so the speed is 30 ÷ 10 = 3 m/s. From 10 s to 20 s the line is flat, so the object has stopped. The last part rises 60 m in 10 s, so the speed is 6 m/s.
Higher tier: on a curved graph the object is accelerating. To find its speed at one moment, draw a tangent, which is a straight line that just touches the curve at that point. The gradient of the tangent is the speed at that moment.
Higher tier: motion in a circle has a constant speed but a changing velocity, because the direction keeps changing. A car going round a roundabout at a steady 10 m/s is still accelerating.
A distance-time graph. The line rises from 0 to 30 metres in 10 seconds, stays flat until 20 seconds, then rises to 90 metres at 30 seconds. The steeper last part shows a faster speed.
Acceleration and velocity-time graphs
Acceleration is how quickly the velocity changes. The bigger the change in a given time, the bigger the acceleration. A sports car that gets to 20 m/s in 5 s has a larger acceleration than a bus that takes 20 s.
acceleration = change in velocity ÷ time taken, or a = Δv ÷ t in symbols. The Greek letter Δ is called delta and just means change in. The unit is metres per second squared, written m/s². A car that goes from 0 to 20 m/s in 5 s has an acceleration of 20 ÷ 5 = 4 m/s². Its speed goes up by 4 m/s every second.
Slowing down is a negative acceleration, also called deceleration. A train going from 24 m/s to rest in 8 s has a change in velocity of −24 m/s. Its acceleration is −24 ÷ 8 = −3 m/s², and the minus sign only tells you it is slowing down.
Near the Earth's surface, any object falling freely has an acceleration of about 9.8 m/s². We often use 10 m/s². Everyday accelerations are smaller. A car pulling away from traffic lights has an acceleration of about 2 m/s², and a hard emergency stop is a deceleration of about 8 m/s². You can write an estimate with the symbol ~, which means about.
A velocity-time graph shows a whole journey on one picture. Time goes along the bottom and velocity goes up the side. A line sloping up means speeding up. A flat line means constant velocity, so the acceleration is zero. A line sloping down means slowing down. The gradient of the line is the acceleration. In a straight line, a speed-time graph looks the same.
Look at the graph. In the first 5 s the line rises by 20 m/s, so the acceleration is 20 ÷ 5 = 4 m/s². From 5 s to 10 s the line is flat, so the car is cruising at 20 m/s. In the last 4 s the line falls from 20 m/s to 0, so the car slows by 20 ÷ 4 = 5 m/s².
Higher tier: the area under a velocity-time graph is the distance travelled. The flat part is 20 m/s × 5 s = 100 m. A sloping part is a triangle, so use ½ × base × height. The whole journey is 50 + 100 + 40 = 190 m. On a curved graph you can count squares.
Higher tier: when the acceleration is constant, v² − u² = 2 × a × s. Here u is the starting speed, v is the final speed and s is the distance travelled. From rest with a = 4 m/s² over 8 m, v² = 2 × 4 × 8 = 64, so v = 8 m/s.
A velocity-time graph. The line rises from 0 to 20 metres per second in 5 seconds, stays flat until 10 seconds, then falls to 0 at 14 seconds.
Newton's second law in use
A car accelerates because the forces on it do not cancel out. There is a resultant force, which is the left-over force once every force has been added up. If the resultant is zero, the acceleration is zero and a moving object carries on at a constant velocity.
Newton's second law links the three ideas. The acceleration of an object is proportional to the resultant force on it, and inversely proportional to its mass. We write a ∝ F and a ∝ 1 ÷ m. As an equation, resultant force = mass × acceleration, or F = m × a. The force is in newtons, the mass is in kilograms and the acceleration is in m/s². You also meet this law in the forces lesson.
Two things follow from it. Double the resultant force and the acceleration doubles. Double the mass and the acceleration halves. You know this from pushing an empty trolley and then a full one.
To find the acceleration, divide: a = F ÷ m. A van of mass 1000 kg with a resultant force of 2500 N has a = 2500 ÷ 1000 = 2.5 m/s². Always use the resultant force, not just the engine force.
Higher tier: inertial mass is a measure of how difficult it is to change the velocity of an object. It is defined as the ratio of force to acceleration, m = F ÷ a. An object with a big inertial mass hardly accelerates when you push it.
You can use the law to estimate the forces in road transport. A typical car has a mass of about 1000 kg. If it speeds up from 0 to 27 m/s, which is about 60 mph, in about 9 s, its acceleration is about 3 m/s². The resultant force is about 1000 × 3 = 3000 N.
You can test the law with a trolley. A hanging weight pulls the trolley along a track, and a light gate measures the acceleration. To change the force but keep the mass the same, move small masses from the trolley to the hanger. The acceleration goes up in proportion to the force. To change the mass, add masses to the trolley while the pulling force stays the same. The acceleration goes down as the mass goes up.
Three bars showing the acceleration in metres per second squared given by the same 4 newton force to masses of 1, 2 and 4 kilograms. The accelerations are 4, 2 and 1.
Thinking, braking and stopping
Imagine you are cycling and a dog runs out in front of you. You need a moment to see it and react, and only then do you squeeze the brakes. A car and its driver work in exactly the same way.
Thinking distance is how far the vehicle travels while the driver reacts. In that time the car keeps moving at full speed, so thinking distance = speed × reaction time. A typical reaction time is between about 0.2 s and 0.9 s. At 20 m/s with a reaction time of 0.7 s, the car goes 20 × 0.7 = 14 m before the brakes are even pressed.
Braking distance is how far the vehicle travels after the brakes go on, until it stops. The brakes use friction to slow the wheels. The friction force does work, which transfers the kinetic energy of the car to the thermal store of the brakes, so they get hot.
The faster the car is going, the bigger the braking force it needs to stop in a certain distance. A bigger braking force gives a bigger deceleration. For the same braking force, a higher speed gives a longer stopping distance.
stopping distance = thinking distance + braking distance. Always add the two parts. A car with a thinking distance of 6 m and a braking distance of 15 m needs 6 + 15 = 21 m to stop.
Real figures show how quickly the distance grows. A typical car needs about 12 m to stop from 20 mph, about 23 m from 30 mph, about 53 m from 50 mph and about 96 m from 70 mph. Speed matters a great deal.
Both distances are measured in metres. A longer stopping distance means a higher chance of hitting something. That is why speed limits exist and why you should keep a good gap behind the car in front.
A flow of four boxes: the driver sees the hazard, the thinking distance while the car keeps moving, the braking distance with the brakes on, and the car stops. The first two distances added make the stopping distance.
What changes the stopping distance
Some things change the thinking distance. Tiredness, alcohol, drugs and distractions such as a phone all make a driver react more slowly. A longer reaction time means a longer thinking distance.
You can measure your own reaction time with a ruler. A partner holds a 30 cm ruler hanging down, with the zero level with the top of your open finger and thumb. They let go without warning and you catch it as fast as you can. The distance the ruler falls shows your reaction time. A short fall means a quick reaction, and a fall of about 20 cm is a reaction time of about 0.2 s.
Repeat the test at least five times and find the mean. Then test again with a distraction, such as talking or looking at a phone. The ruler falls further, so the reaction time is longer. Use the same hand and the same partner each time, so that the test is fair.
Speed changes both distances. At a higher speed the car covers more road in the same reaction time. It also has more kinetic energy that the brakes must remove.
Other things change only the braking distance. Wet or icy roads have less friction, so the tyres grip less. Worn brakes and worn tyres also grip less, and a heavily loaded car has more mass to stop.
Kinetic energy = ½ × mass × speed². Because the speed is squared, doubling the speed makes the kinetic energy four times bigger. With the same braking force, the braking distance is then about four times longer.
Large decelerations are dangerous. A very big braking force can overheat the brakes, and it can make the tyres lose their grip, so the driver loses control. The people in the car also feel a large force. This is why you should brake gently when you can.
Higher tier: you can estimate the braking force. A car of mass 1000 kg slows from 20 m/s to rest in 4 s. The deceleration is 20 ÷ 4 = 5 m/s², so the braking force is about 1000 × 5 = 5000 N.
Terminal velocity
Drag is the force from air or water that pushes against a moving object. In air it is also called air resistance. The faster the object moves, the bigger the drag.
Picture a skydiver jumping from a plane. At the start she is slow, so there is hardly any drag. Her weight pulls her down, so the resultant force is large and she accelerates quickly.
As her speed rises, the drag grows. The resultant force, which is weight minus drag, gets smaller. She still speeds up, but more and more slowly.
Eventually the drag equals her weight. The forces are balanced, so the resultant force is zero and the acceleration is zero. She keeps falling at a constant speed called the terminal velocity.
The same thing happens to any object falling through a fluid, which means a liquid or a gas. A ball bearing sinking in thick oil speeds up and then falls at a steady speed. On a velocity-time graph, the line for a skydiver curves upwards and then levels off at the terminal velocity.
Opening the parachute makes the drag suddenly bigger than the weight, so she slows down. As she slows, the drag falls until it equals the weight again. She now falls at a much lower terminal velocity, which is safe to land at.
A skydiver with an upward arrow labelled drag and a downward arrow labelled weight. The arrows are the same length, so the forces are balanced and the skydiver falls at terminal velocity.
Recap
speed = distance ÷ time, and distance = speed × time. Speed is a scalar and velocity is a vector. On a distance-time graph the gradient is the speed and a flat line means the object is stationary. Typical speeds are walking 1.5 m/s, running 3 m/s, cycling 6 m/s and sound 330 m/s.
acceleration = change in velocity ÷ time, in m/s². A negative value means slowing down. On a velocity-time graph, the gradient is the acceleration and a flat line means constant velocity. Higher tier: the area under the graph is the distance, and v² − u² = 2 × a × s.
resultant force = mass × acceleration. The same force gives a smaller acceleration to a bigger mass. A zero resultant force means constant velocity. Drag grows with speed, so a falling object reaches terminal velocity when the drag equals the weight.
stopping distance = thinking distance + braking distance. Tiredness, alcohol and distractions raise the thinking distance, and a ruler drop test measures reaction time. Wet roads and worn brakes or tyres raise the braking distance, and a higher speed raises both. Very large decelerations can overheat brakes and cause a loss of control.
How this is asked in an exam: you may be asked for a speed from a distance and a time, an acceleration from a speed change and a time, or a force from a mass and an acceleration. Graph questions ask for a gradient. Longer questions ask you to explain terminal velocity. A good answer says the weight is constant, the drag grows with speed, the forces become balanced, the resultant force is zero and so the speed stays constant.
Key points
speed = distance ÷ time. On a distance-time graph, the gradient is the speed.
acceleration = change in velocity ÷ time, in m/s². On a velocity-time graph, the gradient is the acceleration.
resultant force = mass × acceleration, so a bigger mass gives a smaller acceleration.
Terminal velocity is reached when the resultant force is zero, so the drag balances the driving force. For a skydiver falling, that means the drag equals the weight.
stopping distance = thinking distance + braking distance. Reaction times are about 0.2 s to 0.9 s.
Higher tier: the area under a velocity-time graph is the distance. v² − u² = 2 × a × s for constant acceleration.
Worked examples
Example 1
A cyclist rides at a steady 6 m/s for 40 s. Calculate the distance travelled.
Write the equation: distance = speed × time.
Substitute the numbers: distance = 6 × 40.
Work out the answer and add the unit: 240 m.
Answer: 240 m
Common slips
Slip: distance = speed ÷ time To find the distance, multiply the speed by the time. Dividing gives a speed.
Slip: 6 ÷ 40 = 0.15 Multiply 6 by 40. Dividing gives 0.15, which is not a distance.
Slip: 46 m 6 × 40 = 240. The 46 came from adding the two numbers.
Example 2
A motorbike accelerates from 0 to 18 m/s in 6 s. Calculate its acceleration.
Write the equation: acceleration = change in velocity ÷ time.
Substitute the numbers: 18 ÷ 6.
Work out the answer and add the unit: 3 m/s².
Answer: 3 m/s²
Common slips
Slip: acceleration = change in velocity × time Divide the change in velocity by the time. Multiplying does not give an acceleration.
Slip: 24 m/s² 18 ÷ 6 is 3. The 24 came from adding 18 and 6.
Example 3
Higher tier. A car starts from rest and accelerates at 3 m/s² over a distance of 24 m. Calculate its final speed.
Write v² − u² = 2 × a × s. Starting from rest means u = 0.
Substitute the numbers: v² = 2 × 3 × 24 = 144.
Take the square root to find v: v = 12 m/s.
Answer: 12 m/s
Common slips
Slip: v² − u² = a × s Do not forget the 2. The equation is v² − u² = 2 × a × s.
Slip: v² = 2 + 3 + 24 = 29 Multiply 2 × a × s. Adding the numbers is not the method.
Slip: v = 144 m/s 144 is v², not v. Take the square root to get 12 m/s.
Questions and answers
Motion, forces and stopping practice questions for GCSE Combined Science Physics. Try each one, then open the answer. Easy questions come first.
Question 1 Easy
A cyclist speeds up from 0 to 12 m/s in 4 s. Work out the acceleration.
Show answer
Answer: 3 m/s²
a = change in velocity ÷ time
12 ÷ 4 = 3 m/s²
Question 2 Easy
A 4 kg object accelerates at 5 m/s². Work out the resultant force.
Show answer
Answer: 20 N
F = ma
4 × 5 = 20 N
Question 3 Easy
A car has a thinking distance of 12 m and a braking distance of 24 m. Work out the stopping distance.
Show answer
Answer: 36 m
stopping distance = thinking + braking
12 + 24 = 36 m
Question 4 Easy
State what is meant by terminal velocity.
Show answer
Answer: The constant speed reached when the resultant force is zero (for a falling object, drag equals weight).
The forces are balanced.
It is the constant speed reached when drag equals weight.
Question 5 Medium
A resultant force of 600 N acts on a 200 kg scooter. Work out its acceleration.
Show answer
Answer: 3 m/s²
a = F ÷ m
600 ÷ 200 = 3 m/s²
Question 6 Medium
A car slows from 30 m/s to rest in 6 s. Work out the size of the acceleration.
Show answer
Answer: 5 m/s²
a = change in velocity ÷ time
30 ÷ 6 = 5 m/s²
Question 7 Medium
State two factors that increase the thinking distance of a driver.
Show answer
Answer: tiredness and alcohol
Anything that slows the reaction time.
Tiredness and alcohol both increase it.
Question 8 Medium
State two factors that increase the braking distance of a car.
Show answer
Answer: wet roads and worn brakes
Anything that reduces the grip or braking force.
Wet roads and worn brakes both increase it.
Exam-style questions
Longer questions in the style of a written paper, with the marking points. These are Yumio practice questions.
Exam-style question 14 marks
A car of mass 1000 kg accelerates from rest. The resultant force on it is 3000 N.
(a) Work out the acceleration of the car. [2 marks]
(b) Work out the speed of the car after 4 s. [2 marks]
Show answer and marking points
Answer: (a) 3 m/s² (b) 12 m/s
(a) acceleration = force ÷ mass (1)
(a) 3 m/s² (1)
(b) speed = acceleration × time (1)
(b) 12 m/s (1)
Exam-style question 24 marks
A driver sees a hazard and brakes. The thinking distance is 9 m and the braking distance is 24 m.
(a) Work out the stopping distance. [1 mark]
(b) Describe one change that increases the thinking distance. [1 mark]
(c) Describe and explain how a wet road changes the stopping distance. [2 marks]
Show answer and marking points
Answer: (a) 33 m (b) The driver is tired. (c) The braking distance increases because there is less friction between the tyres and the road.
(a) 33 m (1)
(b) tiredness, alcohol, distraction or speed (1)
(c) the braking distance increases (1)
(c) less friction between the tyres and the road (1)
Common mistakes
Using the speed instead of the change in speed
Acceleration = change in velocity ÷ time. Subtract the start speed from the end speed first.
Saying terminal velocity means the skydiver stops
The skydiver keeps falling at a constant speed. It is the acceleration that is zero.
Saying a bigger mass gives a bigger acceleration for the same force
A bigger mass gives a smaller acceleration, because a = F ÷ m.
Leaving out the thinking distance when finding the stopping distance