The equations of motion, also known as SUVAT equations, are used when acceleration, $a$, is constant. They are known as SUVAT equations because they contain the following variables: $s$ - distance, $u$ - initial velocity, $v$ - velocity at time $t$, $a$ - acceleration and $t$ - time. However, each SUVAT equation does not contain all variables so for answering some questions it might be necessary to use one or more of them. The equations are as follows:
\begin{align*} v &= u + at,\\ s & = \left(\frac{u+v}{2}\right)t, \\ v^2 &= u^2 + 2as,\\ s &= ut + \frac{1}{2}at^2,\\ s & = vt - \frac{1}{2}at^2. \end{align*}
The quantities $s$, $u$, $v$ and $a$ are all vector quantities so therefore their sign represents the direction of motion.
You need to ensure that the measurements are in base SI units before substituting them into the formulae. These are:
\begin{align*} \text{time } t &= \text{seconds } \mathrm{s},\\ \text{displacement } s &= \text{metres } \mathrm{m}, \\ \text{velocity } v \text{ or } u & = \text{metres per second } \mathrm{ms^{-1} },\\ \text{acceleration }a &=\text{metres per second per second } \mathrm{ms^{-2} }. \end{align*}
==Example using $v = u + at$==
If a car is moving with an initial speed of $10\mathrm{ms}^{-1}$ and accelerates at $5\mathrm{ms^{-2} }$ for $3 \mathrm{s}$, at what speed is it travelling when it stops accelerating?
From the question we know the values of some of the variables. It is best to write down what we know in a list \begin{align} s &= ?\\ u &= 10\\ v &= ?\\ a &= 5\\ t &= 3 \end{align} From the list of values we know we need to use the equation $v = u + at$ to find $v$. On substitution we obtain: \begin{align} v &= u + at,\\ &=10 + \left(5\times 3\right),\\ &=25 \mathrm{ms^{-1} }. \end{align} The car is travelling at $25\mathrm{ms^{-1} }$ when it stops accelerating.
Note: In these types of questions remember to conclude your calculations with a contextual summary sentence.
Suppose that a ball is dropped out of a plane and it accelerates towards the ground at approximately $9.8\mathrm{ms^{-2} }$. After $4.05 \mathrm{s}$ it hits the ground and is travelling at $40\mathrm{ms^{-1} }$. At what speed was it travelling when it left the plane?
From the question we know that \begin{align} s &= ?\\ u &= ?\\ v &= 40\\ a &= 9.8\\ t &= 4.05 \end{align} We want to find $u$, the initial speed, from the values we have. The equation we have that includes $u$, $v$, $a$ and $t$ is \[v = u + at.\] However, this needs rearranging to make $u$ the subject of the equation. This gives \[u = v - at.\] On substitution of the values we know we obtain \begin{align} u &= v - at,\\ &= 40 - \left(9.8 \times 4.5\right),\\ &=0.31 \mathrm{ms^{-1} }. \end{align} When the ball left the plane it was travelling at $0.31 \mathrm{ms^{-1} }$ towards the ground.
==Example using $s = \left(\frac{u+v}{2}\right)t$==
If a cyclist is moving from $A$ to $B$ in $6$ seconds with an initial velocity at point $A$ of $5\mathrm{ms^{-1} }$ and final velocity at point $B$ of $10\mathrm{ms^{-1} }$, what is the distance in $\mathrm{m}$ from $A$ to $B$?
From the question we know the values of some of the variables. It is best to write down what we know in a list \begin{align} s &= ?\\ u &= 5\\ v &= 10\\ a &= ?\\ t &= 6 \end{align} From the list of values we know we need to use the equation $s = \left(\frac{u+v}{2}\right)t$ to find $s$. On substitution we obtain: \begin{align} s& = \left(\frac{u+v}{2}\right)t,\\ &= \left(\frac{5 + 10}{2}\right) \times 6,\\ &=45 \mathrm{m}. \end{align} The distance from $A$ to $B$ is $45 \mathrm{m}$.
Suppose that a car moves from rest with constant acceleration from a point $A$ to a point $B$. It takes $\frac{2}{3}$ of a minute to get to point $B$ and at this point it is travelling at $60 \mathrm{kmh^{-1} }$. the distance from $A$ to $B$.
Immediately we spot that the final velocity, $v$ and the time, $t$ are not in SI units. So we must convert into SI units using $1 \mathrm{km} = 1000\mathrm{m}$, $1$hour$ = 60 \times 60\mathrm{s} =3600\mathrm{s}$ and $1$ minute $=60 \mathrm{s}$. Therefore for $v$ we have \begin{align} 60 \mathrm{kmh^{-1} } & = 60 \times 1000 \div 3600 \mathrm{ms^{-1} }, \\ & = \frac{50}{3} \mathrm{ms^{-1} }. \end{align} For $t$ we have \begin{align} \frac{2}{3} \mathrm{minute} & = \frac{2}{3} \times 60\mathrm{s}, \\ & = 40\mathrm{s}. \end{align} We can now write down what we know in a list \begin{align} s &= ?\\ u &= 0\\ v &= \frac{50}{3} \\ a &= ?\\ t &= 40 \end{align} Notice $u=0$ because the car begins at rest. From the list of values we know we need to use the equation $s = \left(\frac{u+v}{2}\right)t$ to find $s$. On substitution we obtain: \begin{align} s& = \left(\frac{u+v}{2}\right)t,\\ &= \left(\frac{0 + 5/3}{2}\right) \times 40,\\ &=\frac{100}{3} \mathrm{m}. \end{align} The distance from $A$ to $B$ is $\frac{100}{3}\mathrm{m}$.
==Example using $v^2 = u^2 + 2as$==
A car is travelling at $22 \mathrm{ms^{-1} }$ when it suddenly brakes. The car travels $40\mathrm{m}$ whilst braking and then comes to a stop. What was the acceleration of the car whilst it was braking?
As with previous questions it is best to list what we know first \begin{align} s &= 40\\ u &= 22\\ v &= 0\\ a &= ?\\ t &= ? \end{align} We know that the final velocity must be $v=0$ since the question states that that the car comes to a stop.
The equation which uses $s$, $u$, $v$ and $a$ is \[v^2 = u^2 + 2as.\] This can be rearranged to make $a$ the subject: \[a = \dfrac{v^2 - u^2}{2s}.\]
We can now substitute in the values of the variables that we know: \begin{align} a &= \dfrac{v^2 - u^2}{2s},\\ &=\dfrac{0^2 - 22^2}{2\times 40},\\ &=-6.05 \mathrm{ms^{-2} }. \end{align}
The car accelerates at $-6.05\mathrm{ms^{-2} }$. This is equivalent to decelerating at a rate of $6.05\mathrm{ms^{-2} }$.
== Example using $s = ut + \frac{1}{2}at^2$==
If a cyclist accelerates from $2 \mathrm{ms^{-1} }$ at a rate of $0.05 \mathrm{ms^{-2} }$, how long will it take for them to travel $1.5 \mathrm{km}$?
The information in the question tells us that \begin{align} s &= 1500\\ u &= 2\\ v &= ?\\ a &=0.05\\ t &=? \end{align} The equation we need to use is \[s = ut + \frac{1}{2}at^2.\] Substituting in the values we know gives: \begin{align} s &= ut + \frac{1}{2}at^2,\\ 1500 &= 2\times t + \frac{1}{2} \times 0.05 \times t^2,\\ 1500 &= 2t + 0.025 t^2.\\ \end{align} We must first rearrange the equation into the form “(quadratic)$= 0$” to give \[t^2 + 80t - 60000 = 0.\] Then using the quadratic formula gives \begin{align} t=\frac{-80\pm \sqrt{80^2-4\times 1 \times (-60000)}~}{2\times 1},\\ \ t = 208.19\mathrm{s} \text{ or } t = -288.19\mathrm{s} \text{ (both to }2\text{ d.p.)} \end{align}
However, time cannot be negative. Therefore $t=208.19 \mathrm{s}$.
The cyclist will take $208.19\mathrm{s}$ or approximately $3.5 \mathrm{minutes}$ to travel $1.5\mathrm{km}$.
Suppose that a ball rolls on a straight horizontal line from point $A$, at time $t=0$, to point $B$ with constant deceleration $2 \mathrm{ms^{-2} }$. At point $A$ the ball is travelling at a velocity $10\mathrm{ms^{-1} }$ and the distance to point $B$ is $17\mathrm{m}$. the times that the ball rolls through point $B$, the velocities when the ball rolls through $B$ and the time the ball returns to $A$.
The information in the question tells us that \begin{align} s &= 17\\ u &= 10\\ v &= ?\\ a &=-2\\ t &=? \end{align} Note that $a$ is negative as we are decelerating. The equation we need to use to find $t$ is \[s = ut + \frac{1}{2}at^2.\] Substituting in the values we know gives \begin{align} s &= ut + \frac{1}{2}at^2,\\ 17 &= \left(10\times t\right) - \left(\frac{1}{2} \times 2 \times t^2\right),\\ 17 &= 10t - t^2.\\ \end{align} We must first rearrange the equation into the form “(quadratic)$= 0$” to give \[t^2 - 10t +17 = 0.\] Then using the quadratic formula gives \begin{align} t=\frac{10\pm \sqrt{10^2-\left(4\times 1 \times 17\right)}~}{2\times 1},\ \ t &= 7.828\mathrm{s} \text{ or } t = 2.172\mathrm{s} \text{ (both to }3\text{ d.p.)} \end{align}
The ball rolls through point $B$ after $2.172\mathrm{s}$, then again after $7.828\mathrm{s}$. This is because the ball is rolling with constant deceleration so will eventually change direction.
We now want the velocities that the ball passes through $B$ with. So we will use the equation $v=u+at$ for each value of $t$. When $t=2.172$ we obtain \begin{align} v &=u+at,\\ &= 10 - \left(2 \times 2.172\right),\\ &=5.656 \mathrm{ms^{-1} } \end{align} and when $t=7.828$ we obtain \begin{align} v &= u +at, \\ &=10 - \left(2 \times 7.828\right), \\ &=-5.656 \mathrm{ms^{-1} }. \end{align} This means the ball passes through point $B$ with velocity $5.656\mathrm{ms^{-1} }$ in the direction $A$ to $B$ when $t=2.172$. Then at $t=7.828$ the ball passes through the point $B$ again with velocity $-5.656 \mathrm{ms^{-1} }$ which means it passes with velocity $5.656\mathrm{ms^{-1} }$ in the opposite direction, $B$ to $A$.
To calculate the time it takes for the ball to return to $A$ we use the equation $s = ut + \frac{1}{2}at^2$ but set $s=0$ as the displacement of the ball from it's starting position is now zero. So we obtain \begin{align} 0 & = 10t - \left(\frac{1}{2} \times 2t^2\right), \\ & = t \left(10 - t\right). \\ \ t = 0 \text{ or } t = 10 \mathrm{s}. \end{align} When $t=0$ this represents the starting position of the ball. So the time taken for the ball to travel from $A$ through $B$ and back is $10\mathrm{s}$. Note: There will be some point $C$ which is the furthest point from $A$ where the ball changes direction, at this point the velocity of the ball will be $v=0$ for an instant.
==Example using $s = vt - \frac{1}{2}at^2$==
A truck drives along a straight road with constant acceleration passing a corner shop first and then a warehouse $48 \mathrm{m}$ later. The truck passes the warehouse $20 \mathrm{s}$ after the shop with velocity $10 \mathrm{ms^{-1} }$. What is the trucks acceleration?
The information in the question tells us that \begin{align} s &= 48\\ u &= ?\\ v &= 10\\ a &=?\\ t &=20 \end{align} The equation we need to use to find $a$ is \[s = vt - \frac{1}{2}at^2.\] Substituting in the values we know gives \begin{align} s &= vt - \frac{1}{2}at^2,\\ 48 &= \left(10\times 20\right) - \left(\frac{1}{2} \times a \times 20^2\right),\\ 48 &= 200 - 200a.\\ \end{align} Rearranging for $a$ this becomes \begin{align} 200a & = 200-48, \\ a & = \frac{200-48}{200}, \\ & = 0.76 \mathrm{ms^{-2} }. \end{align}
The acceleration of the truck is $0.76\mathrm{ms^{-2} }$.
Try our Numbas test on equations of motion: Equations of Motion