SUVAT equations

 This adventure is on the SUVAT equations, fundamental to year 1 of maths and physics A level. As I enter my last year of education I am going to start using my blogging time to go over content from the first year of my courses as I think it is crucial to doing well in 8 months time. Any way let me begin and take you through what the SUVAT equations are.

The SUVAT equations define motion at a constant acceleration. What does this mean, it means that if an object start from rest and accelerates to a velocity eg 40 km/h, we will assume it accelerates at a uniform rate as if it wasn't a uniform rate of acceleration then the SUVAT equations can not be used. You must remember this in order to grasp the way you use the SUVAT equations.

Now that you know the condition needed to fulfil the use of SUVAT equations, now you must learn what each letter means.

S = distance

U = initial velocity 

V = final velocity

A = acceleration 

T = time

In order to use the SUVAT equations we need at least 3 variables to find one. We must use 1 of the 5 possible equations.

$\begin{array}{cc}v& =u+at,\\ s& =\left(\frac{u+v}{2}\right)t,\\ v^2& =u^2+2as,\\ s& =ut+\frac{1}{2}a{t}^2,\\ s& =vt-\frac{1}{2}a{t}^2.\end{array}$ 

We can eliminate one of the variables, usually it is because one of either the initial or final velocity is 0. It could be a vehicle coming to rest or a vehicle starting from rest. Let's do a question!

If a car is moving with an initial speed of $10{m\mathrm{/s}}^{}$ and accelerates at $5m{\mathrm{/s\; for}}^{}$$3s$, at what speed is it travelling when it stops accelerating?

U = 10,      A = 5,    T = 3

using V = U + AT

10 + (5 x 3) = 25 m/s.

This is one of the simpler type of SUVAT equation, however there are some that are more challenging.

For example:

A ball is thrown up into the air and lands in the same place it was thrown. Find the maximum height reached given that the initial velocity is 5m/s.

U = 5,     A = -g (-9.8),    V = 0

we use the equation V^2 = U^2 + 2as and rearrange it into s = V^2 - U^2 / 2a. Now we can sub in the values:

0 - 25 / -19.6 = 1.275. 

So the highest point that the ball reached 1.28m.

These are the two types of SUVAT questions I have come across apart from projectiles, which are entirely different types of SUVAT's. Although the standard type of projectile questions are not too difficult, they can be extremely difficult if you have never done the specific type of question it is. That adventure is soon to come, but soon with me is about 2 months.