Equations of motion

In a nutshell

There are four equations that can be used to calculate quantities involving motion at a constant acceleration. These equations are called the suvat equations.

Equations

DESCRIPTION	EQUATION
Equation $1$ without $s$	$v =u + at$
Equation $2$ without $v$	$s = ut + \dfrac{1}{2} at^2$
Equation $3$ without $a$	$s = \dfrac{1}{2} (u+v)t$
Equation $4$ without $t$	$v^2 = u^2 + 2as$

Variables

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$initial \ velocity$	$u$	$ms^{-1}$	$ms^{-1}$
$final \ velocity$	$v$	$ms^{-1}$	$ms^{-1}$
$displacement$	$s$	$m$	$m$
$time$	$t$	$s$	$s$
$acceleration$	$a$	$ms^{-2}$	$ms^{-2}$

The suvat equations

With the four equations of motion you will see in this lesson, you will be able to calculate quantities involving motion in a straight line at constant acceleration. These equations are known as the suvat equations, due to this being an acronym for the variables above. They can all be derived from the velocity-time graph.

The final velocity, $v$ , is equal to the initial velocity $u$ , plus the gradient $a$ multiplied by the change in time $t$ . This gives equation $1$ :

Equation 1

$v = u +a t$

The area under the graph is the displacement $s$ of the object. This will be split into a triangle and rectangle, with areas $\dfrac{1}{2} (v-u)t$ and $ut$ respectively. By re-arranging equation (1), you get $(v – u) = at$ , which must then be substituted into the area of a triangle. This gives $\dfrac{1}{2} \times at \times t$ , adding this to the area of the rectangle $ut$ gives the total area $s$ . Therefore, equation of motion $2$ , without the final velocity $v$ is:

Equation 2

$s = ut + \dfrac{1}{2}at^2$

The acceleration is equal to the gradient of the line on a velocity-time graph. This is also equal to the change in velocity, the final velocity $v$ minus initial velocity $u$ divided by the time taken $t$ , this gives $a = \dfrac{(v-u)}{t}$ . This must be rearranged to get $at = (v-u)$ , then this can be substituted into the previous equation to obtain the final form of the equation. This gives equation $3$ without $a$ :

Equation 3

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

Finally rearranging the equation for acceleration $a = \dfrac{(v-u)}{t}$ to make time the subject, $t = \dfrac{(v-u)}{a}$ . Taking equation without $a$ , $s = \dfrac{1}{2} (u+v)t$ , substituting in the previous equation, you get equation $4$ without $t$ :

Equation 4

$v^2 = u^2 + 2as$

Conditions for the equations

The suvat equations can be used for both horizontal and vertical motion. The conditions for vertical motion are:

When the maximum height is reached, $v=0$ .
Acceleration vertically is due to gravity, which on Earth, $g = 9.81 \ ms^{-2}$ .
Downwards is usually taken as the positive direction, as gravity is an attractive force.
The negative direction must have a negative displacement and velocity in this direction. If downwards is negative, anything going up must be positive.

Note: The suvat equations can only be used for constant acceleration. If the acceleration changes to another constant value, then this motion can be calculated separately as a second period of motion.

Example 1

A man runs along a straight horizontal road with a constant acceleration of $2 \ ms^{-2}$ . The man was running at an initial speed of $3 \ ms^{-1}$ , and during this time covers a distance of $650 \ m$ . Calculate the final speed the man was running at.

First state the variables given in the question:

$s=650\,m \newline u=3 \, ms^{-1} \newline v=? \newline a=2\, ms^{-2} \newline t= \boxtimes$

An equation without $t$ is given by:

$v^2 = u^2 + 2as$

Finally, substitute in and solve:

$v^2 = 3^2 + 2 \times 2 \times 650 \newline \\[0.1in] v^2 = 2609\newline \\[0.1in] v = 51 \ ms^{-1}$

The final speed the man was running at is $\underline{v = 51 \ ms^{-1}}$

Example 2

In Bilbao, divers dive from a rock $27 \ m$  above sea level. Assuming that air resistance is negligible, calculate how long the dive lasts.

First, state the variables given in the question:

$s= 27 \ m, \newline u= 0ms^{-1}, \newline v= \boxtimes \newline a= 9.81 \ ms^{-1} \newline t= ?$

An equation without $v$ is given by:

$s = ut + \dfrac{1}{2}at^2\newline \\[0.1 in] \, but \ u=0, \ \ \ so: s= \dfrac{1}{2}at^2$

Finally, rearrange and substitute in:

$t^2 = \dfrac{2s}{a}\newline \\[0.1in] t^2 = \dfrac{2 \times 27}{9.81}\newline \\[0.1in] t^2 = 5.5\newline \\[0.1in] t = 2.3 \ s$

The dive lasted for $\underline{2.3 \ s}$ .