Acceleration and the equations of motion

Three equations handle almost every Year 11 motion calculation, as long as the acceleration is constant. You need to be able to use them, choose between them, and derive them from a velocity–time graph.

Acceleration

Acceleration is the rate of change of velocity. It is a vector, measured in m s⁻².

$$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}$$

Here $u$ is the initial velocity and $v$ is the final velocity. An acceleration of 3 m s⁻² means the velocity changes by 3 m s⁻¹ every second. Instantaneous acceleration is the gradient of the tangent to a velocity–time graph, just as instantaneous velocity is the gradient of a position–time tangent.

The sign matters. If the positive direction is forwards, a car braking has negative acceleration. But "negative acceleration" doesn't always mean slowing down: a car reversing and speeding up has negative velocity and negative acceleration. An object speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs.

Deriving the equations from a velocity–time graph

1. From the gradient. The gradient of a $v$–$t$ graph is the acceleration: $a = \dfrac{v - u}{t}$. Rearranging:

$$v = u + at$$

2. From the area. The area under a $v$–$t$ graph is the displacement (velocity × time = displacement for each thin strip). Split the area into a rectangle and a triangle:

$$s = \underbrace{ut}_{\text{rectangle}} + \underbrace{\tfrac12 (v - u)t}_{\text{triangle}}$$

Since $v - u = at$, the triangle is $\tfrac12 (at)t$, so

$$s = ut + \tfrac12 at^2$$

The same trapezium also gives $s = \tfrac12(u + v)t$: the average of the initial and final velocities, times the time. That works only for constant acceleration.

3. Eliminating $t$. From $v = u + at$, $t = \dfrac{v - u}{a}$. Substitute into $s = \tfrac12(u + v)t$:

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

All three assume constant acceleration. If the acceleration changes, go back to the graph and use gradients and areas.

Choosing the right equation

The method markers want: list what you know with signs (s, u, v, a, t), mark what you need, pick the equation that leaves out the remaining one, substitute, then solve.

Worked example: braking distance

A car travelling at 16.7 m s⁻¹ (60 km/h) brakes with a constant deceleration of 7.0 m s⁻². Find its braking distance, and the braking distance from 33.3 m s⁻¹ (120 km/h).

$u = 16.7$, $v = 0$, $a = -7.0$, $s = ?$; $t$ not needed.

list (forwards positive)

$v^2 = u^2 + 2as \Rightarrow 0 = 16.7^2 + 2(-7.0)s$.

equation without t

$s = \dfrac{278.9}{14.0} = 19.9$ m.

solve

$s = \dfrac{33.3^2}{14.0} = 79.2$ m. Since $s \propto u^2$, doubling the speed quadruples the braking distance.

double the speed

Worked example: two-object catch-up

A car passes a stationary police car at a constant 20 m s⁻¹. At that instant the police car accelerates from rest at 4.0 m s⁻². When and where does it catch the car, and how fast is it going then?

Car: $s_C = 20t$. Police: $s_P = \tfrac12(4.0)t^2 = 2.0t^2$.

write each position

$2.0t^2 = 20t \Rightarrow t(2.0t - 20) = 0 \Rightarrow t = 10$ s (the other root, $t = 0$, is the start).

catch-up means same position

$s = 20 \times 10 = 200$ m from the start.

where

$v = u + at = 0 + 4.0 \times 10 = 40$ m s⁻¹, twice the car's speed.

police speed

Vertical motion under gravity

Near Earth's surface, with air resistance ignored, every object accelerates downwards at $g = 9.8$ m s⁻². This holds on the way up, at the top and on the way down. With up as positive, $a = -9.8$ m s⁻² for the whole flight.

Worked example: throwing a ball up from a cliff

A ball is thrown straight up at 15 m s⁻¹ from the edge of a cliff 20 m above the sea. Find (a) its maximum height above the cliff, (b) the time to reach the sea and (c) its velocity as it hits the water.

Up positive, origin at the cliff top: $u = +15$, $a = -9.8$.

sign convention

$0 = 15^2 + 2(-9.8)s \Rightarrow s = \dfrac{225}{19.6} = 11.5$ m above the cliff.

(a) top: v = 0

$-20 = 15t - 4.9t^2 \Rightarrow 4.9t^2 - 15t - 20 = 0$, so $t = \dfrac{15 + \sqrt{225 + 392}}{9.8} = 4.07$ s (reject the negative root).

(b) sea is at s = −20 m

$v = 15 - 9.8 \times 4.07 = -24.8$ m s⁻¹, so 24.8 m s⁻¹ downwards. Check: $v^2 = 15^2 + 2(-9.8)(-20) = 617$, $|v| = 24.8$.

(c) velocity

InteractiveCatch-up and cliff-throw lab

Try this

• In Catch the speeding car with the defaults (20 m/s, 4.0 m/s², no delay), check the worked example: caught after 10 s at 200 m, police car at 40 m/s.
• Add a 2 s reaction delay. The catch now happens at 13.71 s and 274.2 m, at 46.8 m/s. Look at the v–t graph: catch-up is where the two shaded areas are equal, not where the lines cross.
• What is happening when the two v–t lines cross?
• Switch to Throw a ball up from a cliff with 15 m/s and 20 m. Check: top after 1.53 s, 11.48 m above the cliff, and it reaches the sea after 4.07 s at −24.84 m/s.
• Set the cliff height to 0. How does the landing speed compare with the launch speed?

Test yourself

1. A train slows uniformly from 25 m s⁻¹ to rest in 10 s. Its stopping distance is

   62.5 m250 m125 m500 m

2. A ball is thrown vertically upwards. At the highest point, its velocity and acceleration are

   0 and 9.8 m s⁻² down0 and 09.8 m s⁻¹ down and 00 and 9.8 m s⁻² up

3. A car accelerates from rest at 2.5 m s⁻² for 8.0 s. The distance it covers is

   20 m40 m160 m80 m

4. The speed of a car is tripled. Assuming the same braking deceleration, its braking distance is multiplied by

   396√3

Try again

Label the velocity–time graph used in the derivations.

CheckStart again

Which equation would you use first?

Find the speed of a car 6 s after it starts accelerating at 3 m/s²Find how long a ball takes to reach the top of its flightFind how far a dropped stone falls in 2.0 sFind when a police car catches a speeding carFind a car's braking distance from 100 km/hFind how high a ball rises when thrown up at 12 m/s

CheckStart again

The gradient of a velocity–time graph is the , and the area under it is the . The equations of motion are only valid when the acceleration is . To find a final velocity without knowing the time, use $v^2 = u^2 + 2as$. An object slows down when its velocity and acceleration have signs.

CheckStart again

5 marks

Using a velocity–time graph for an object accelerating uniformly from $u$ to $v$ in time $t$, derive $s = ut + \tfrac12at^2$. Then use the equations of motion to derive $v^2 = u^2 + 2as$.

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