The SUVAT Equations and their derivation

The SUVAT Equations describe motion in a given direction when ACCELERATION IS CONSTANT. The SUVAT Equations that are given in the Formula Booklet are:

  1. \hspace{10pt}V=U+AT
  2. \hspace{10pt}S=\left(\frac{U+V}{2}\right)T
  3. \hspace{10pt}V^2=U^2+2AS
  4. \hspace{10pt}S=UT+\frac{1}{2}AT^2
  5. \hspace{10pt}S=VT-\frac{1}{2}AT^2
VariableDescriptionSI unit
Uinitial velocitym/s 
Vfinal velocitym/s
Ttotal times

Derivation of the SUVAT Equations

In general, for motion in a straight line with constant acceleration:

\text{acceleration} = \frac{\text{change in velocity}}{\text{change in time}}\hspace{3pt}\text{or}\hspace{3pt}A=\frac{V-U}{T}

where V is the final velocity, U is the initial velocity and T is the total time taken. Rearranging gives the equation in an alternative form:


This equation is one of the SUVAT equations. They are named so since they involve displacement (S), initial velocity (U), final velocity (V), acceleration (A) and time (T) for motion in a straight line with constant acceleration. Note that in order to use the SUVAT equations, we must define a reference point with regards to the displacement. We must also specify the direction of positive and negative speed. Speed cannot be negative but acceleration can. If an object is slowing down rather than speeding up, acceleration is negative.

The second SUVAT equation S=\frac{1}{2}(U+V)T comes from the fact that acceleration is constant. In this case, \frac{1}{2}(U+V) is the average speed throughout the duration of travel. Multiplying this by T will give the total distance. This is because distance is speed multiplied by time when acceleration is constant.

The first two SUVAT equations can be used to derive the remaining SUVAT equations:

  • SUVAT Equation 1 can be rearranged to make T the subject so that T=\frac{V-U}{A} which can be substituted into equation 2:
    S=\left(\frac{U+V}{2}\right)\left(\frac{V-U}{A}\right) and rearranged gives V^2=U^2+2AS
  • Substitute the expression for V in SUVAT Equation 1 directly into SUVAT Equation 2:
  • SUVAT Equation 1 can be rearranged to make U the subject so that U=V-AT. Substitute this into equation 4 to give


See the Examples below for different applications of the SUVAT equations. When attempting examples for yourself, make sure that the dimensions are consistent. In other words, you should ensure that you are using the same units for all measurements. Accordingly, this may require a conversion as in Examples 3 and 4. You should also see Motion under Gravity for more examples using the SUVAT equations.


A stone is dropped from the edge of a very tall cliff. Find the speed of the stone after 4 seconds of falling. How far does the stone travel in this time?

We identify this as a SUVAT equations question since the stone, treated as a particle, is acting under constant acceleration – gravity. Picking downwards as the positive direction, we can set A=9.8 m/s/s. Since the particle is dropped, it has an initial speed of U=0 m/s. The stone falls for 4 seconds and so we set T=4 s and we are looking to find the final velocity V (final for the purposes of this question). The displacement is not given and is not required and so can be ignored. To summarise this information:

  • S (not needed to find speed)
  • U=0
  • V= ?
  • A=9.8
  • T=4

and we note that all the units are consistent. The first SUVAT equation above relates U, V, A and T and so we choose this one to find V:

V=U+AT\hspace{10pt}\Longrightarrow\hspace{10pt}V=0+9.8\times 4=39.2

The velocity of the stone after 4 seconds of falling is 39.2 m/s.

Now that we have found the velocity we can update the SUVAT characteristics:

  • S= ?
  • U=0
  • V=39.2
  • A=9.8
  • T=4

We can now choose any SUVAT equation that involves S to find the displacement of the stone. The second SUVAT equation is the simplest:

S=\left(\frac{U+V}{2}\right)T=\left(\frac{0+39.2}{2}\right)\times 4=78.4

The stone travels 78.4m in the first 4 seconds of falling.

Note that many SUVAT equation questions ask you to find one missing feature before moving on to find another.

A particle starts moving from the origin with speed 4m/s and moves along the x-axis with a constant deceleration of 2m/s/s. Find the times at which the particle passes the point at which x=3.

We begin by identifying the SUVAT characteristics in turn and then the SUVAT equation to use becomes apparent. The displacement from the origin is S=3, the initial speed is U=4, V is not given nor required, given that we are moving initially in the positive x-direction we set A=-2 and T is what we are trying to find. To summarise:

  • S=3
  • U=4
  • V (not needed)
  • A=-2
  • T=?

and again we note that the units are consistent. The fourth SUVAT equation above relates S, U, A and T:

S=UT+\frac{1}{2}AT^2\hspace{10pt}\Longrightarrow\hspace{10pt}3=4T+\frac{1}{2}\times -2\times T^2

which rearranged and factorised gives


implying T=1 or T=3. Hence, the particle passes the point where x=3 after 1 seconds and again after 3 seconds. This is because the particle starts with positive speed and is decelerating – eventually the particle slows down to 0, turns around and heads back towards the origin.

A devoted stargazer notices what he thinks is a UFO (what is a UFO?) in the night sky. He manages to clock the speed of the object and the distance travelled at 550m/s and 2km respectively, approximately 5 seconds after noticing it. Assuming the object is travelling with constant acceleration in a straight line, find an estimate for its acceleration.

Summarising the information and making the units consistent:

  • S=2000
  • U (not needed)
  • V=550
  • A=?
  • T=5

The final SUVAT equation above relates S, V, A and T:

S=VT-\frac{1}{2}AT^2\hspace{10pt}\Longrightarrow\hspace{10pt}2000=550\times 5-\frac{1}{2}A\times 5^2

Rearranging gives:

A=\frac{2000-2750}{-\frac{1}{2}\times 25}=60. Hence, the UFO has an acceleration of 60 m/s/s.

A car is travelling along a straight road with lampposts with constant acceleration.

a) It takes the car 5 seconds to travel between the first and second lampposts. Given that the first and second lampposts are 30 metres apart and the car passes the second lamppost with a speed of 27km/h, find the speed of the car when it passed the first lamppost.

b) From the second lamppost, the car then speeds up with a new constant acceleration of 0.5m/s/s. The car passes the third lamppost with speed 36km/h. Find the distance between the second and third lampposts in metres.

a) We first note that in this question the units are not consistent and so we change 30m to \frac{30}{1000}=\frac{3}{100}km and 5s to \frac{5}{3600}=\frac{1}{720}hr. We do this because the speeds are given in km/h and the question is asking for a new speed. We summarise the information as follows:

  • S=\frac{3}{100}
  • U=?
  • V=27
  • A (not needed)
  • T=5

The second SUVAT Equation above relates S, U, V and T:

S=\left(\frac{U+V}{2}\right)T\hspace{10pt}\Longrightarrow\hspace{10pt}\frac{3}{100}=\left(\frac{U+V}{2}\right)T=\left(\frac{U+27}{2}\right)\times \frac{1}{720}

Rearranging gives U=\frac{3}{100}\times 720\times 2 - 27=16.2km/h.

b) Note that the final velocity from the first part of the question becomes the initial velocity for the second part. We change the speeds in km/h to m/s. Firstly, 27 km/h is the same as 27000 metres in 3600 seconds or 7.5 m/s. Secondly, 36 km/h is the same as 36000 metres in 3600 seconds or 10m/s. To summarise:

  • S=?
  • U=7.5
  • V=10
  • A=0.5
  • T (not needed)

The SUVAT Equation above that relates S, U, V and A is…

V^2=U^2+2AS\hspace{10pt}\Longrightarrow\hspace{10pt}10^2=7.5^2+2\times 0.5\times S

Solving gives S=43.75 and the distance between the second and third lampposts is 43.75 metres.


This video will show you how to tackle a basic SUVAT equations question by first setting up the equations then showing you how to select which one to use. This particular question asks students how to find how long it takes for a ball thrown in the air takes to reach the ground under certain conditions. Initial speed is given and gravity is set to 10m/s^2. There were two exam questions on this paper using SUVAT equations – SUVAT Equation 2 will show you how to answer the second.

This video will show you how to tackle a more involved SUVAT equations question. There are several parts to the question – some use SUVAT in the several parts, some use a velocity-time graph and the final asks the student to suggest an improvement to the model. There were two exam questions on this paper using SUVAT equations – SUVAT Equation 1 will show you how to answer the other one which was Question 6 on this paper. This question is a little more basic and sets up the SUVAT equations before showing you how to select which one to use to find the time taken for a ball thrown in the air to reach the ground.

AS Mechanics Kinematics

AS Mechanics Kinematics