The SUVAT equations describe motion in a given direction when ACCELERATION IS CONSTANT.

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

\text{acceleration} = \frac{\text{change in velocity}}{\text{change in time}}\hspace{30pt}\text{or}\hspace{30pt}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:

V=U+AT

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.


The full set of equations that you should commit to memory are given by:

  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

where

Variable/Constant Description SI unit
S displacement m (metres)
U initial velocity m/s (metres per second)
V final velocity m/s (metres per second)
A acceleration m/s/s (metres per second per second)
T total time s (seconds)

 

Note that in order to use these 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.


Example 1

A ball starts rolling down a hill from rest with a constant acceleration of 6.7 m/s/s. Find the velocity of the ball after 4 seconds. How far has the ball travelled in that time?




Example 2

A cyclist is in a race and 100 metres from the finish he decides to accelerate his speed. The cyclist maintains a constant acceleration of 0.4m/s/s. If the cyclist crosses the finish line with a speed of 17m/s, how fast was he cycling when he started to accelerate?




Example 3

A particle is moving along a line parallel with the x-axis. The particle is observed at the origin with a speed of 12 micrometres per second. Point A is 160 micrometres from the origin. If the particle decelerates at a constant rate of 0.4 micrometres per second per second, find the times that the particle passes point A.


See Motion under Gravity for more examples using 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 SI unit for all measurements. Accordingly, this may require a conversion.


Derivation of the SUVAT equations

We saw SUVAT equation 1 above. 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 3 – 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
  • SUVAT EQUATION 4 – substitute the expression for V in SUVAT equation 1 directly into SUVAT equation 2: S=\left(\frac{U+U+AT}{2}\right)T=...=UT+\frac{1}{2}AT^2
  • SUVAT EQUATION 5 – SUVAT equation 1 can be rearranged to make U the subject so that U=V-AT. Substitute this into equation 4 to give S=\left(V-AT\right)T+\frac{1}{2}AT^2=...=VT-\frac{1}{2}AT^2

Example

Consider the expression for final velocity V V=U+AT where U is initial velocity, A is constant acceleration and T is time. Given also that displacement S is given by S=\frac{1}{2}(U+V)T, show that V^2=U^2+2AS.

Rearrange V=U+AT to make T the subject:

V=U+AT\hspace{3pt} \Rightarrow\hspace{3pt} V-U=AT\hspace{3pt}\Rightarrow\hspace{3pt}\left(\frac{V-U}{A}\right)=T

Substituting this expression for T into the given expression for S gives:

S=\frac{1}{2}(U+V)\times\left(\frac{V-U}{A}\right)=\frac{1}{2A}(U+V)(V-U)

Multiplying both sides by 2A and expanding the brackets gives:

2AS=U^2-V^2\hspace{4pt}\Rightarrow\hspace{4pt}V^2=U^2-2AS