A Level Maths: SUVAT
SUVAT – Introduction
In this article we are going to introduce an area of A Level Mechanics which is known as SUVAT.
SUVAT is not a standalone topic and will also appear in other areas of mechanics such as questions involving Newton’s Laws.
What is important is that when it comes to questions involving SUVAT you are doing a little diagram to help you visualise the question, regardless of how simple you think the question is.
Label the information that you have and it is also a good idea to learn the formulas rather than relying on finding them in the formula booklet. This is because, if you know them, then you are more likely to use the correct SUVAT formula when doing A Level maths.
In this article we will look at articles that are travelling both horizontally and vertically under gravity.
SUVAT - Uniform acceleration formula
When the motion of a body is being considered then the letters u, v, a, t and s are generally used. And they represent the following:
u – initial velocity
v – final velocity
a – acceleration
t – time
s – displacement/distance
There are five important formulas that need to be remembered.
\begin{gathered} v=u+a t \\ v^2=u^2+2 a s \\ s=\frac{(u+v) t}{2} \\ s=u t+\frac{1}{2} a t^2 \\ s=v t-\frac{1}{2} a t^2 \end{gathered}SUVAT - Example
Solution
When it comes to questions involving SUVAT it is always a very good idea to write down the information that you have got.
a) a=\frac{2}{3}, t=12, v=25
In order to find u we can use the SUVAT equation v=u+a t
25=u+\frac{2}{3}(12) \therefore u=17 \mathrm{~ms}^{-1}b) To find the distance we can use the SUVAT equation s=u t+\frac{1}{2} a t^2
s=17(12)+\frac{1}{2}\left(\frac{2}{3}\right) 12^2 \therefore s=252 m
Example
A coin is dropped from rest at the top of a building which is of height 12m and travels in a straight line with constant acceleration of 10 ms-2. Find the time it takes to reach the ground and the speed of impact.
Solution
Writing the information we have:
u-0, a-10, s-12, t-t, v-vIn order to find time we can use the SUVAT equation s=u t+\frac{1}{2} a t^2
12=0(12)+\frac{1}{2}(10) t^2 \therefore t=1.55 \text { seconds }In order to find the speed upon impact we can use v^2=u^2+2 a s
v^2=2(10)(12) \therefore v=15.5 \mathrm{~ms}^{-1}Example
You are using your stopwatch facility to measure the time between lamp posts on a car journey. As the car speeds up two consecutive times are 1.2 seconds and 1 second. Later you find out that the lamp posts are 30m apart.
a) What is the acceleration of the car and its speed at the first lamp post?
b) Taking the same value for acceleration, what is the time taken for the car to travel the 30m before the first lamp post?
Solution
a) This is clearly a more complicated question involving SUVAT and a diagram to visualise what is happening will be useful.
Consider the motion from A to B and we want u and a.
\text { Using } s=u t+\frac{1}{2} a t^2 \text { then } \mathbf{3 0}=1.2 u+0.72 aIf we consider the motion from B to C then the final velocity from A to B will be the initial velocity from B to C and this is something that we do not have and introducing it involves another unknown.
So we will consider the motion from A to C where we have
s=60, u=u, a=a, t=2.2Using s=u t+\frac{1}{2} a t^2 \text { then } 60=2.2 u+2.42 a
The two equations found involve two unknowns and so we can solve them simultaneously to give a=4.55 \mathrm{~ms}^{-2} \text { and } u=22.3 \mathrm{~ms}^{-1}
b) To answer this part of the question suppose we have the following:
Here s=30, v=22.3, a=4.55, t=t
Using s=v t-\frac{1}{2} a t^2 we have:
\begin{aligned} &30=22.3 t-2.275 t^2\\ &2.275 t^2-22.3 t+30=0 \end{aligned}Using the quadratic formula to find t we obtain the values of 1.61 and 8.19. Now both times are positive but based on previous time intervals the most sensible answer will be 1.61 seconds.
Free fall under gravity
With the uniform acceleration formula that we have met so far, these can also be used when considering a body that is falling freely under gravity. In such cases we take the acceleration of the body to be 9.8 \mathrm{~ms}^{-2} \text {. }
If the body is travelling upwards then there will be a negative acceleration.
Example
A brick is thrown vertically downwards from the top of a building and has an initial velocity of 1.5 \mathrm{~ms}^{-1}. If the height of the building is 19 \frac{2}{7} \mathrm{~m} find:
a) The velocity with which the brick hits the ground
b) The time taken for the brick to fall
Solution
a) Here we have the following information: u=1.5, a=9.8, s=19 \frac{2}{7}, v=v
b) Here we can use v^2=u^2+2 a s:
\begin{gathered} v^2=1.5^2+2(9.8)\left(19 \frac{2}{7}\right) \\ v^2=380.25 \therefore v=19.5 \mathrm{~ms}^{-1} \end{gathered}b) Here we can use the following SUVAT equation: \begin{gathered} v-u+a t \\ 19.5=1.5+9.8 t \\ t=1.84 \mathrm{~s} \end{gathered}
Example
A particle is projected vertically upwards with a velocity of 34.3 ms-1. Find how long after initial projection the particle is at a height of 49m above its point of projection
a) For the first time
b) For the second time
Solution
Here we are given:
u=34.3, a=-9.8, s=49, t=tSo using the SUVAT equation s=u t+\frac{1}{2} a t^2 we have:
\begin{gathered} 49=34.3 t-4.9 t^2 \\ t^2-7 t+10=0 \\ (t-5)(t-2)=0 \end{gathered}So the particle is 49m above the initial point of projection
a) 2 seconds, for the first time
b) 5 seconds, for the second time
SUVAT questions can get tricky. Keep your work clear and neat and label everything clearly.