🚀 Starting from rest
The initial speed is zero, so u = 0 and u2 = 0.
The equation shrinks to v2 = 2 × a × s.
Look out for the words "from rest", "stationary" or "dropped".
Movement and Position » Final Speed, Acceleration and Distance
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 1.10
- How to use v2 = u2 + (2 × a × s) to link speed, acceleration and distance
- How to spot the questions that need this equation: the ones with no time in them
- How to rearrange the equation to find the acceleration or the distance moved
- How to deal with objects that start from rest, brake to rest or fall
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Unlock This CourseAn equation with no time in it
So far, every motion calculation has needed a time. But many questions never mention one. A question might tell you how far an object moves while it speeds up, and ask how fast it ends up going. For these you need a new relationship:
(final speed)2 = (initial speed)2 + (2 × acceleration × distance moved)
v2 = u2 + (2 × a × s)
| Symbol | Quantity | Unit |
|---|---|---|
| v | final speed | m/s |
| u | initial speed | m/s |
| a | acceleration | m/s2 |
| s | distance moved | m |
Take care with the letter s. In this equation it stands for the distance moved. It does not stand for speed or for seconds.
Good news: this equation is given on the formulae page of the exam paper, so you do not have to memorise it. You do have to know when to pick it and how to use it without slips.
The equation only works when the acceleration is constant for the whole distance. Exam questions tell you this, or give you one value for the acceleration.
Finding the final speed
Notice that the equation gives you v2, not v. There are always two jobs to do: square the initial speed at the start, and take the square root at the end.
Worked example 1: a quad bike
A quad bike is moving at 5 m/s. It accelerates at 2 m/s2 while it travels 36 m along a farm track. Calculate its final speed.
List what you know: u = 5 m/s, a = 2 m/s2, s = 36 m. You want v.
v2 = u2 + (2 × a × s)
v2 = 52 + (2 × 2 × 36)
v2 = 25 + 144 = 169
v = √169
v = 13 m/s
Work out the bracket as one number (144 here) before you add it to u2. Then press the square root key. If you stop at 169, you have found v2 and you lose the final mark.
Two special cases: from rest and to rest
🛑 Braking to rest
The final speed is zero, so v = 0 and v2 = 0.
The equation becomes 0 = u2 + (2 × a × s).
The acceleration comes out negative, because the object is slowing down.
Worked example 2: from rest
A rollercoaster car starts from rest. It accelerates at 3 m/s2 along 24 m of straight track. Calculate its final speed.
u = 0 m/s, a = 3 m/s2, s = 24 m
v2 = u2 + (2 × a × s)
v2 = 02 + (2 × 3 × 24)
v2 = 144
v = √144 = 12 m/s
Rearranging to find the acceleration or the distance
Sometimes you know both speeds and need a or s. Rearrange in two steps.
- Subtract u2 from both sides. This gives v2 − u2 = 2 × a × s.
- Divide by the two things you do not want. To find a, divide by 2 × s. To find s, divide by 2 × a.
a = (v2 − u2) ÷ (2 × s)
s = (v2 − u2) ÷ (2 × a)
Worked example 3: braking to rest
An ambulance is travelling at 20 m/s. The driver brakes and it comes to rest in a distance of 50 m. Calculate its acceleration.
u = 20 m/s, v = 0 m/s, s = 50 m
a = (v2 − u2) ÷ (2 × s)
a = (02 − 202) ÷ (2 × 50)
a = (0 − 400) ÷ 100
a = −4 m/s2
The minus sign shows that the ambulance is slowing down. Its deceleration is 4 m/s2.
Worked example 4: finding the distance
A bobsleigh speeds up from 10 m/s to 20 m/s with an acceleration of 2.5 m/s2. Calculate the distance it moves while it does this.
u = 10 m/s, v = 20 m/s, a = 2.5 m/s2
s = (v2 − u2) ÷ (2 × a)
s = (202 − 102) ÷ (2 × 2.5)
s = (400 − 100) ÷ 5
s = 300 ÷ 5 = 60 m
Falling objects
The same equation works for an object that falls straight down. Near the Earth's surface a falling object speeds up with an acceleration of 10 m/s2, as long as air resistance is small enough to ignore. Use a = 10 m/s2 unless a question gives you another value. The distance moved, s, is the height it falls.
An object that is dropped starts from rest, so u = 0.
Worked example 5: a diver
A diver steps off a platform and falls 5 m to the water. Calculate her speed as she reaches the water. Ignore air resistance.
u = 0 m/s, a = 10 m/s2, s = 5 m
v2 = u2 + (2 × a × s)
v2 = 02 + (2 × 10 × 5) = 100
v = √100 = 10 m/s
Worked example 6: how high is the cliff?
A pebble is dropped from the top of a sea cliff. It hits the water at 30 m/s. Calculate the height of the cliff. Ignore air resistance.
u = 0 m/s, v = 30 m/s, a = 10 m/s2
s = (v2 − u2) ÷ (2 × a)
s = (302 − 02) ÷ (2 × 10)
s = 900 ÷ 20 = 45 m
What air resistance does to a falling object is covered in the lesson Falling Objects and Terminal Velocity.
Which equation should you choose?
Write down the quantities the question gives you and the one it asks for. Then check the list:
- A time is given or asked for, and the question is only about speeds and acceleration (no distance): use the equation from the lesson Acceleration.
- A distance is given or asked for, and there is no time: use v2 = u2 + (2 × a × s).
Common mistakes
- Forgetting the square root. The equation gives v2. Square root it to get v.
- Forgetting to square u. In example 1, u2 is 25, not 5.
- Missing out the 2. The bracket is 2 × a × s, not a × s.
- Subtracting before squaring. Square each speed first, then subtract. In example 4, 400 − 100 = 300. Working out (20 − 10)2 gives 100, which is wrong.
- Losing the minus sign. When an object slows down, a is negative. If you are given a deceleration of 4 m/s2, put a = −4 into the equation.
- Mixing units. Speeds must be in m/s and distances in m. Change 0.3 km to 300 m before you start.
Exam-style question
A racing car leaves the pit lane at a speed of 15 m/s. It then accelerates at a constant 5 m/s2 while it travels 80 m.
(a) Explain why the relationship v2 = u2 + (2 × a × s) is suitable for finding the final speed of the car. [1 mark]
(b) Calculate the final speed of the car. [3 marks]
(c) Later, the car is travelling at 60 m/s. The driver brakes and the car comes to rest in a distance of 150 m. Calculate the acceleration of the car while it brakes. [3 marks]
(d) In the garage, a wheel nut is dropped from rest and falls 1.25 m to the floor. Calculate the speed of the wheel nut as it reaches the floor. Use acceleration = 10 m/s2. [2 marks]
Model answer
(a) The distance moved is known but the time taken is not, and this relationship does not need the time. (1)
(b) v2 = 152 + (2 × 5 × 80) (1)
v2 = 225 + 800 = 1025 (1)
v = √1025 = 32 m/s (1)
(c) 02 = 602 + (2 × a × 150) (1)
a = −3600 ÷ 300 (1)
a = −12 m/s2 (1)
(d) v2 = 02 + (2 × 10 × 1.25) = 25 (1)
v = √25 = 5 m/s (1)
Exam tip
In part (b) the calculator shows 32.0156. The numbers in the question have one or two significant figures, so round your answer to 32 m/s, but only at the very end. In part (c) the word "brakes" tells you the answer must be negative. If your acceleration comes out positive, you have swapped u and v.