Equations of motion

AS · 13 min

When acceleration is constant, five quantities describe a straight-line motion completely, and four equations link them. These "suvat" equations are the workhorse of AS mechanics: they appear in free fall, braking, projectiles, collisions and energy questions. You must be able to derive them from the definitions of velocity and acceleration, use them fluently with signs, and describe an experiment to measure the acceleration of free fall.

The five quantities

Symbol	Quantity	Unit
$s$	displacement	$\text{m}$
$u$	initial velocity	$\text{m s}^{-1}$
$v$	final velocity	$\text{m s}^{-1}$
$a$	acceleration (constant)	$\text{m s}^{-2}$
$t$	time taken	$\text{s}$

All except $t$ are vectors, so in one dimension each carries a sign relative to a chosen positive direction.

Key result

For uniform acceleration in a straight line:

v = u + at

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

s = ut + \tfrac{1}{2}at^2

v^2 = u^2 + 2as

Each equation leaves out one of the five quantities: $s$ , $a$ , $v$ and $t$ respectively. The data sheet gives $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$ ; recall the other two.

Deriving the equations

The syllabus requires you to derive these equations from the definitions. There are two routes and you should know both: algebra from the definitions, and areas on a velocity–time graph.

Derivation from the definitions

Equation 1. Acceleration is the rate of change of velocity. If it is constant,

a = \frac{v - u}{t} \quad\Rightarrow\quad v = u + at

Equation 2. When acceleration is uniform, velocity changes steadily from $u$ to $v$ , so the average velocity is $\tfrac{1}{2}(u + v)$ . Displacement $=$ average velocity $\times$ time:

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

Equation 3. Substitute $v = u + at$ into equation 2:

s = \tfrac{1}{2}(u + u + at)t = \tfrac{1}{2}(2u + at)t = ut + \tfrac{1}{2}at^2

Equation 4. From equation 1, $t = \dfrac{v - u}{a}$ . Substitute into equation 2:

s = \tfrac{1}{2}(u + v)\frac{(v - u)}{a} = \frac{v^2 - u^2}{2a} \quad\Rightarrow\quad v^2 = u^2 + 2as

Derivation from a velocity–time graph

For uniform acceleration the velocity–time graph is a straight line from $(0, u)$ to $(t, v)$ . Its gradient is $a$ , so $v = u + at$ .

The displacement is the area under the line, a trapezium with parallel sides $u$ and $v$ and width $t$ :

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

Equivalently, split it into a rectangle of area $ut$ and a triangle of base $t$ and height $v - u = at$ :

s = ut + \tfrac{1}{2}t(at) = ut + \tfrac{1}{2}at^2

Equation 4 then follows by eliminating $t$ as above.

(0, 3) -- (5, 11) (0, 3) -- (5, 3) (5, 0) -- (5, 11) fill 0 5 y = 3 + 1.6 x

The shaded trapezium under the line from $u = 3$ to $v = 11$ splits into a rectangle ( $ut$ ) below the horizontal line and a triangle ( $\tfrac{1}{2}at^2$ ) above it.

Watch out

The equations are valid only when the acceleration is constant (uniform). They cannot be used for a car whose engine force varies, for a falling object with significant air resistance, or across a collision. Between stages with different accelerations, apply them to each stage separately, using the final velocity of one stage as the initial velocity of the next.

Using the equations

Method

Solving a suvat problem

Draw a quick sketch and choose a positive direction. State it.
List $s$ , $u$ , $v$ , $a$ , $t$ . Fill in the three you know (with signs) and mark the one you want.
Choose the equation that contains your three knowns and the unknown (it leaves out the fifth).
Substitute, solve, and give the answer with a unit and a sign or direction.
Check it makes sense (a falling object speeds up; a braking car's distance is positive).

Words hide data. "From rest" means $u = 0$ . "Comes to rest" or "stops" means $v = 0$ . "At the highest point" of a vertical throw means $v = 0$ . "Dropped" means $u = 0$ . "Free fall" or "neglecting air resistance" means $a = \pm 9.81\ \text{m s}^{-2}$ , with the sign set by your convention.

A car accelerating

A car accelerates uniformly from $8.0\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ over a distance of $84\ \text{m}$ . Calculate the acceleration and the time taken.

Solution

$s = 84\ \text{m}$ , $u = 8.0\ \text{m s}^{-1}$ , $v = 20\ \text{m s}^{-1}$ , $a = ?$ , $t = ?$ .

No $t$ , so use $v^2 = u^2 + 2as$ :

20^2 = 8.0^2 + 2a(84) \quad\Rightarrow\quad a = \frac{400 - 64}{168} = 2.0\ \text{m s}^{-2}

Then $v = u + at$ : $t = \dfrac{20 - 8.0}{2.0} = 6.0\ \text{s}$ .

Check with $s = \tfrac{1}{2}(u + v)t = \tfrac{1}{2}(28)(6.0) = 84\ \text{m}$ .

Free fall

Near the Earth's surface, any object falling with negligible air resistance has the same constant acceleration, the acceleration of free fall, $g = 9.81\ \text{m s}^{-2}$ downwards (given on the data sheet). It does not depend on the mass of the object.

A stone dropped from a cliff

A stone is dropped from the top of a cliff $45\ \text{m}$ high. Neglecting air resistance, find the time it takes to reach the sea and its speed on impact.

Solution

Take downwards as positive: $s = 45\ \text{m}$ , $u = 0$ , $a = 9.81\ \text{m s}^{-2}$ .

s = ut + \tfrac{1}{2}at^2 \quad\Rightarrow\quad 45 = \tfrac{1}{2}(9.81)t^2 \quad\Rightarrow\quad t = \sqrt{\frac{90}{9.81}} = 3.03\ \text{s}

v^2 = u^2 + 2as = 2(9.81)(45) = 882.9 \quad\Rightarrow\quad v = 29.7\ \text{m s}^{-1}

A ball thrown upwards from a height

A ball is thrown vertically upwards at $12\ \text{m s}^{-1}$ from a point $1.5\ \text{m}$ above the ground. Neglecting air resistance, calculate (a) the maximum height above the ground and (b) the time from release until the ball hits the ground.

Solution

Take upwards as positive, so $a = -9.81\ \text{m s}^{-2}$ .

(a) At the top $v = 0$ . With $u = 12$ :

0 = 12^2 + 2(-9.81)s \quad\Rightarrow\quad s = \frac{144}{19.62} = 7.34\ \text{m}

Height above the ground $= 7.34 + 1.5 = 8.84\ \text{m} \approx 8.8\ \text{m}$ .

(b) The ground is $1.5\ \text{m}$ below the release point, so $s = -1.5\ \text{m}$ for the whole flight:

-1.5 = 12t - 4.905t^2 \quad\Rightarrow\quad 4.905t^2 - 12t - 1.5 = 0

t = \frac{12 \pm \sqrt{144 + 4(4.905)(1.5)}}{2(4.905)} = \frac{12 \pm 13.17}{9.81}

The positive root is $t = 2.57\ \text{s}$ . (The negative root, $-0.12\ \text{s}$ , refers to a time before the throw and is rejected.)

Tip

Treating the whole up-and-down flight as one stage works because the acceleration is the same throughout. You do not need to split at the top unless the question asks about the top. The signed displacement $s = -1.5\ \text{m}$ takes care of the fact that the ball ends below where it started.

Multi-stage problems

Stopping distance

A driver travelling at $25\ \text{m s}^{-1}$ sees a hazard. Her reaction time is $0.70\ \text{s}$ , after which the car decelerates uniformly at $6.5\ \text{m s}^{-2}$ . Calculate the total stopping distance.

Solution

Stage 1, thinking distance. During the reaction time the car moves at constant velocity:

s_1 = 25 \times 0.70 = 17.5\ \text{m}

Stage 2, braking distance. $u = 25$ , $v = 0$ , $a = -6.5$ :

0 = 25^2 + 2(-6.5)s_2 \quad\Rightarrow\quad s_2 = \frac{625}{13} = 48.1\ \text{m}

Total $= 17.5 + 48.1 = 65.6\ \text{m} \approx 66\ \text{m}$ .

Doubling the speed doubles the thinking distance but quadruples the braking distance, because $s \propto u^2$ when $v = 0$ .

Catching up

A truck passes a parked car at a constant $15\ \text{m s}^{-1}$ . At that instant the car starts from rest with a constant acceleration of $2.5\ \text{m s}^{-2}$ in the same direction. Find the time taken for the car to catch the truck, the distance travelled, and the car's speed at that moment.

Solution

When the car catches the truck, both have the same displacement in the same time $t$ .

Truck: $s = 15t$ . Car: $s = \tfrac{1}{2}(2.5)t^2 = 1.25t^2$ .

15t = 1.25t^2 \quad\Rightarrow\quad t(1.25t - 15) = 0 \quad\Rightarrow\quad t = 12\ \text{s}

( $t = 0$ is the starting moment.) Distance $= 15 \times 12 = 180\ \text{m}$ . Car's speed $= u + at = 2.5 \times 12 = 30\ \text{m s}^{-1}$ , exactly twice the truck's, as it must be: both have the same average velocity, and the car's average is half its final velocity.

Falling past a window

A ball is dropped from rest above a window. It takes $0.25\ \text{s}$ to fall past the window, which is $2.0\ \text{m}$ tall. Neglecting air resistance, find how far above the top of the window it was released.

Solution

Take downwards as positive. First find the speed $u$ at the top of the window, using the motion past the window: $s = 2.0\ \text{m}$ , $t = 0.25\ \text{s}$ , $a = 9.81\ \text{m s}^{-2}$ .

2.0 = u(0.25) + \tfrac{1}{2}(9.81)(0.25)^2 = 0.25u + 0.307 \quad\Rightarrow\quad u = \frac{2.0 - 0.307}{0.25} = 6.77\ \text{m s}^{-1}

Now the fall from rest to the top of the window: initial velocity $0$ , final velocity $6.77\ \text{m s}^{-1}$ .

6.77^2 = 0 + 2(9.81)h \quad\Rightarrow\quad h = \frac{45.9}{19.62} = 2.34\ \text{m}

Determining the acceleration of free fall

The syllabus asks you to describe an experiment to determine $g$ using a falling object.

Electromagnet and trapdoor method: switching off the electromagnet releases the ball and starts the timer; the ball hitting the trapdoor stops it. The height h is measured from the bottom of the ball to the trapdoor.

Determining g with a falling ball

Apparatus. A steel ball held by an electromagnet above a trapdoor switch (or a ball falling through two light gates connected to a data logger); an electronic timer; a metre rule; a stand and clamp.

Method.

Set the height $h$ from the bottom of the ball to the trapdoor, using a metre rule held vertically (check with a set square). The ball's bottom is the part that hits the trapdoor.
Switch off the electromagnet. This releases the ball and starts the timer; the trapdoor opening stops it. Record the fall time $t$ .
Repeat for the same $h$ at least three times and average.
Repeat for at least five different heights spread over the full available range (for example $0.30\ \text{m}$ to $1.10\ \text{m}$ ).

Analysis. The ball starts from rest, so $s = ut + \tfrac{1}{2}at^2$ becomes

h = \tfrac{1}{2}g\,t^2

Plot $h$ (on the $y$ -axis) against $t^2$ . The graph should be a straight line through the origin with gradient $g/2$ , so $g = 2 \times \text{gradient}$ .

Variables. Independent: $h$ . Dependent: $t$ . Controlled: the same ball, released from rest each time, negligible air resistance (use a small dense ball).

Sources of error and improvements.

The electromagnet does not release the ball instantly as the current falls (residual magnetism): a systematic delay. Using a graph helps: a constant delay shows up in the intercept rather than spoiling the gradient (strictly it affects $t$ rather than $h$ , so plotting $t$ against $\sqrt{h}$ makes the delay the intercept).
Measuring $h$ to the wrong part of the ball: always measure to the bottom of the ball.
Air resistance: small for a dense ball over short distances; it reduces the measured $g$ .
Using a hand-held stop-watch would give a reaction-time error comparable with the fall time (about $0.4\ \text{s}$ ): that is why an electronic timer or light gates are used.

With light gates, the ball (or a card of known length) passes through two gates a measured distance apart. The data logger gives the speed at each gate (length of card $\div$ time to pass) and the time between gates, so $g = (v_2 - v_1)/t$ ; or it gives $v$ at several heights for a graph of $v^2$ against $h$ , gradient $2g$ .

Analysing free-fall data

A student obtains these results with the electromagnet method.

$h$ / m	$t$ / s	$t^2$ / s $^2$
0.300	0.248	0.0615
0.500	0.319	0.1018
0.700	0.378	0.1429
0.900	0.429	0.1840
1.100	0.473	0.2237

The line of best fit on a graph of $h$ against $t^2$ passes through $(0.0500,\ 0.244)$ and $(0.2200,\ 1.080)$ . Determine $g$ .

Solution

y = 4.917 x - 0.0021 (0.0615, 0.300) (0.1018, 0.500) (0.1429, 0.700) (0.1840, 0.900) (0.2237, 1.100)

\text{gradient} = \frac{1.080 - 0.244}{0.2200 - 0.0500} = \frac{0.836}{0.170} = 4.92\ \text{m s}^{-2}

Since $h = \tfrac{1}{2}g t^2$ , the gradient is $g/2$ :

g = 2 \times 4.92 = 9.84\ \text{m s}^{-2} \approx 9.8\ \text{m s}^{-2}

The line passes very close to the origin, as expected if there is no systematic timing delay.

Exam tip

"Derive" the equations of motion means starting from the definitions $a = (v - u)/t$ and average velocity $= (u + v)/2$ , or from areas under a velocity–time graph, and showing every algebraic step.
Always state your sign convention, and give $a$ the correct sign (for example $-9.81\ \text{m s}^{-2}$ if up is positive). Mixing conventions within one problem is the most common source of lost marks.
When a quadratic in $t$ gives two roots, state which one you take and why.
For the free-fall experiment, the marks are for: how $h$ and $t$ are measured, repeating and varying $h$ , the graph to plot, and how $g$ is obtained from the gradient. "Use the equation $s = \tfrac{1}{2}gt^2$ " with no graph is usually given only partial credit.

Summary

The four equations apply only to uniform acceleration in a straight line.
Derive them from $a = (v - u)/t$ and $s = \tfrac{1}{2}(u + v)t$ , or from areas on a $v$ – $t$ graph.
List $s, u, v, a, t$ ; choose the equation that omits the quantity you neither know nor need.
Choose a positive direction and keep it; in free fall $a = \pm 9.81\ \text{m s}^{-2}$ .
A whole vertical up-and-down flight can be one stage, with signed displacement.
Multi-stage problems: the final velocity of one stage is the initial velocity of the next.
Determine $g$ by plotting $h$ against $t^2$ for a ball falling from rest: gradient $= g/2$ .

Practice

Question

Derive $v^2 = u^2 + 2as$ from $v = u + at$ and $s = \tfrac{1}{2}(u + v)t$ .
A train accelerates uniformly from rest at $0.40\ \text{m s}^{-2}$ . How far does it travel in $30\ \text{s}$ , and what is its final speed?
A cyclist travelling at $9.0\ \text{m s}^{-1}$ brakes uniformly and stops in $15\ \text{m}$ . Calculate the deceleration and the braking time.
A stone is thrown vertically downwards at $5.0\ \text{m s}^{-1}$ from a bridge $20\ \text{m}$ above a river. Find its speed on hitting the water and the time taken.
A ball is thrown vertically upwards and returns to the thrower's hand $2.4\ \text{s}$ later. Find its initial speed and the maximum height reached.
A car travelling at $30\ \text{m s}^{-1}$ needs $75\ \text{m}$ to stop under maximum braking. What stopping distance would it need at $15\ \text{m s}^{-1}$ with the same deceleration (ignore thinking distance)?
An object starts with velocity $4.0\ \text{m s}^{-1}$ and accelerates uniformly at $3.0\ \text{m s}^{-2}$ . Find the distance it travels during the fourth second of its motion (from $t = 3.0\ \text{s}$ to $t = 4.0\ \text{s}$ ).
In a free-fall experiment, a student plots $h$ against $t^2$ and obtains a gradient of $4.75\ \text{m s}^{-2}$ . Calculate $g$ . Suggest one reason why the value is lower than $9.81\ \text{m s}^{-2}$ .
A hot-air balloon is rising vertically at a steady $4.0\ \text{m s}^{-1}$ . When it is $30\ \text{m}$ above the ground, a sandbag is released from it. Calculate (a) the maximum height of the sandbag above the ground and (b) the time it takes to reach the ground.
Two balls are released from the same height of $20\ \text{m}$ . Ball A is dropped; ball B is released $0.50\ \text{s}$ later with a downward velocity $u$ . Both reach the ground at the same instant. Find $u$ .

Answers

From $v = u + at$ , $t = (v - u)/a$ . Then $s = \tfrac{1}{2}(u + v)(v - u)/a = (v^2 - u^2)/(2a)$ , so $v^2 = u^2 + 2as$ .
$s = \tfrac{1}{2}(0.40)(30)^2 = 180\ \text{m}$ ; $v = 0.40 \times 30 = 12\ \text{m s}^{-1}$ .
$0 = 9.0^2 + 2a(15)$ , so $a = -2.7\ \text{m s}^{-2}$ (deceleration $2.7\ \text{m s}^{-2}$ ). $t = (0 - 9.0)/(-2.7) = 3.3\ \text{s}$ .
Down positive: $v^2 = 5.0^2 + 2(9.81)(20) = 417.4$ , so $v = 20.4\ \text{m s}^{-1}$ . $t = (20.4 - 5.0)/9.81 = 1.57\ \text{s}$ .
Time to the top $= 1.2\ \text{s}$ , so $u = 9.81 \times 1.2 = 11.8\ \text{m s}^{-1}$ . Height $= u^2/(2g) = 11.77^2/19.62 = 7.06\ \text{m}$ .
Braking distance $\propto u^2$ , so halving the speed quarters the distance: $75/4 = 18.75\ \text{m} \approx 19\ \text{m}$ .
$s(4.0) = 4.0(4.0) + \tfrac{1}{2}(3.0)(16) = 40\ \text{m}$ ; $s(3.0) = 4.0(3.0) + \tfrac{1}{2}(3.0)(9) = 25.5\ \text{m}$ . Distance in the fourth second $= 14.5\ \text{m}$ .
$g = 2 \times 4.75 = 9.50\ \text{m s}^{-2}$ . Any one reason with its effect: the electromagnet releases the ball slightly after the timer starts, so every measured time is too long; or air resistance reduces the ball's acceleration.
Up positive. The sandbag starts with $u = +4.0\ \text{m s}^{-1}$ (it shares the balloon's velocity). (a) Rise $= 4.0^2/(2 \times 9.81) = 0.82\ \text{m}$ , so maximum height $= 30.8\ \text{m}$ . (b) $-30 = 4.0t - 4.905t^2$ , so $4.905t^2 - 4.0t - 30 = 0$ and $t = \dfrac{4.0 + \sqrt{16 + 588.6}}{9.81} = 2.91\ \text{s}$ .
Ball A: $20 = 4.905t^2$ , $t = 2.019\ \text{s}$ . Ball B has $2.019 - 0.50 = 1.519\ \text{s}$ : $20 = 1.519u + 4.905(1.519)^2 = 1.519u + 11.32$ , so $u = 5.7\ \text{m s}^{-1}$ .

The five quantities

Deriving the equations

Using the equations

Free fall

Multi-stage problems

Determining the acceleration of free fall

Summary

Practice

Console