Revision:Kinematics - Equations of Motion for Constant Acceleration

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Revision:Kinematics - Equations of Motion for Constant Acceleration

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1 The 'SUVAT' Equations of Motion
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The 'SUVAT' Equations of Motion

When acceleration is constant, there are four very useful equations that will help you in solving just about any kinematics problem you want. At times it will not always be obvious which one (or ones) to use to solve the problem most efficiently, but working through the problem slowly will usually lead to the correct result. You may or may not be given the equations on the formula sheet, but even if you are - it'll be far more beneficial for you to just learn them. If anything, it saves time!

The Equations

They are called the 'suvat' equations because the quantities s, u, v, a and t are used in the equations, with four of the symbols used in each equation.

= displacement (measured in metres)
= initial velocity (measured in metres per second, ms^-1)
= final velocity (also measured in ms^-1)
= acceleration (measured in metres per second per second, ms^-2)
= time (measured in seconds, s)

Here are the equations:

v = u + at

$s = \frac{1}{2} (u+v)t$
$s = ut + \frac{1}{2} at^2$
v^2 = u^2 + 2as

It will also be useful for you to know how these equations are derived.

Deriving the equations of motion

1. v = u + at

Acceleration is defined as the rate of change of velocity with respect to time. This can be written as
$a=\frac{v-u}{t}$
at = v - u
v = u + at

2. $s = \frac{1}{2}(u+v)t$

Velocity is the rate of change of displacement with respect to time, from here it follows that displacement can be found from the product of average velocity and time. Average velocity is found by adding the final and initial velocities and dividing by 2 i.e. $\frac{u+v}{2}$ . You can then multiply by the time to get .

It is from these two equations that the 3rd and 4th equations can be derived.

3. $s = ut + \frac{1}{2}at^2$

We know from equation 2 that $s = \frac{1}{2}(u+v)t$ and we now want to find another way of expressing s, but this time in terms of u, a and t rather than u, v and t.

However, we also know from equation 1 that v = u + at , and we can substitute this on to equation 2:

$s = \frac{1}{2} (u + v)t$
$s = \frac{1}{2} (u + u + at)t$
$s = \frac{1}{2} (2u + at)t$
$s = \frac{1}{2} (2ut + at^2)$
Which simplifies to give:
$s = ut + \frac{1}{2}at^2$

4. v^2 = u^2 + 2as

Going back to our first equation, we can rearrange this to give $t = \frac{v-u}{a}$ , which then eliminates the t in equation 2, $s = \frac{1}{2}(u+v)t$ . This then gives:

$s = \frac{1}{2} (u+v) \frac{v-u}{a}$
$s = \frac {(v+u)(v-u)}{2a}$
2as = (v+u)(v-u) and spotting the RHS as the difference of two squares gives
2as = v^2-u^2 and therefore
v^2 = u^2 + 2as

Reminder

It is important to bear in mind that these equations can only be used for CONSTANT ACCELERATION ONLY. When acceleration is not constant, these equations do not work. For variable acceleration, either graphical methods or calculus would be needed.

Furthermore, these equations can only be used for motion in a straight line or one-dimensional motion.

Thus these equations are known as the equations of rectilinear motion. Rectilinear motion is one-dimensional motion with uniform acceleration.