Uniform Rectilinear Motion

Lessons

Senior 1 · Physics

Uniform Rectilinear Motion

5 lessons

Lesson 3 of 5

Acceleration

22 min

Leaving the traffic lights

The lights turn green at a junction in Kigali. A moto pulls away, and its speed, read every second, is:

Time $t$ (s)	0	1	2	3	4
Velocity $v$ ( $\mathrm{m\,s^{-1}}$ )	0	3	6	9	12

Every second the velocity rises by the same $3\,\mathrm{m\,s^{-1}}$ . A bus pulling away beside it might gain only $1\,\mathrm{m\,s^{-1}}$ each second. The moto is not just fast: it is gaining speed quickly. The quantity that measures how quickly velocity changes is acceleration.

Definition

Acceleration ( $a$ ) is the rate of change of velocity: the change in velocity divided by the time taken.

a = \frac{v - u}{t}

where $u$ is the initial velocity, $v$ the final velocity and $t$ the time. Acceleration is a vector. SI unit: metre per second squared, $\mathrm{m\,s^{-2}}$ (metres per second, gained every second).

A negative acceleration, when an object slows down, is called a deceleration or retardation. If the velocity changes by the same amount every second, the acceleration is uniform.

Experiment: a ball rolling down a ramp

Raise one end of a smooth plank about 2 m long on a few books, to make a gentle slope.
One pupil claps once every second, steadily. Practise with a watch until the claps are regular.
Hold a ball at the top and release it on a clap, without pushing it.
At each following clap, a partner puts a chalk mark beside the ball. Repeat three times and use the average position of each mark.
Measure the distance from the start to each mark.

Figure 3.1: Positions of the ball at each clap. The gap covered in each second grows by the same 0.20 m every second.

Time (s)	0	1	2	3	4
Distance from the start (m)	0	0.10	0.40	0.90	1.60
Distance covered in that second (m)		0.10	0.30	0.50	0.70

The ball's average speed in each second is 0.10, 0.30, 0.50 and $0.70\,\mathrm{m\,s^{-1}}$ : it rises by $0.20\,\mathrm{m\,s^{-1}}$ every second. The ball has a uniform acceleration of $0.20\,\mathrm{m\,s^{-2}}$ .

Important

Safety. Keep the slope low and put a bag or a box at the bottom to stop the ball. Check that the plank cannot slide off the books.

What acceleration means

Rearranging the definition gives the final velocity after accelerating for a time $t$ :

v = u + a t

The sign of $a$ tells you whether the object speeds up or slows down, when the motion is in the positive direction:

$a > 0$ : the velocity increases (speeding up);
$a = 0$ : the velocity is constant;
$a < 0$ : the velocity decreases (slowing down, a deceleration).

Figure 3.2: Motion diagrams. Each dot is the position after an equal interval of time, and each arrow shows the velocity: the longer the arrow, the faster the car.

Worked example 1: a moto pulling away

The moto in the opening goes from rest to $12\,\mathrm{m\,s^{-1}}$ in 4.0 s. Find its acceleration.

$u = 0$ , $v = 12\,\mathrm{m\,s^{-1}}$ , $t = 4.0\,\mathrm{s}$ :

a = \frac{v - u}{t} = \frac{12\,\mathrm{m\,s^{-1}} - 0}{4.0\,\mathrm{s}} = 3.0\,\mathrm{m\,s^{-2}}

Worked example 2: a bus braking

A bus travelling at $15\,\mathrm{m\,s^{-1}}$ brakes and stops in 5.0 s. Find its acceleration.

a = \frac{v - u}{t} = \frac{0 - 15\,\mathrm{m\,s^{-1}}}{5.0\,\mathrm{s}} = -3.0\,\mathrm{m\,s^{-2}}

The minus sign shows a deceleration of $3.0\,\mathrm{m\,s^{-2}}$ : the bus loses $3\,\mathrm{m\,s^{-1}}$ of speed every second.

Worked example 3: finding the final velocity

A cyclist moving at $2.0\,\mathrm{m\,s^{-1}}$ accelerates at $0.50\,\mathrm{m\,s^{-2}}$ for 6.0 s down a gentle hill. What is her final velocity?

v = u + at = 2.0\,\mathrm{m\,s^{-1}} + 0.50\,\mathrm{m\,s^{-2}} \times 6.0\,\mathrm{s} = 2.0 + 3.0 = 5.0\,\mathrm{m\,s^{-1}}

Warning

Common mistakes. (1) Confusing speed with acceleration: a bus cruising at a steady $60\,\mathrm{km\,h^{-1}}$ on a straight road is fast but has ZERO acceleration. (2) Mixing units: change $\mathrm{km\,h^{-1}}$ to $\mathrm{m\,s^{-1}}$ before you use $a = \frac{v-u}{t}$ . (3) Writing the unit as $\mathrm{m\,s^{-1}}$ : acceleration is in $\mathrm{m\,s^{-2}}$ .

Tip

Read $3\,\mathrm{m\,s^{-2}}$ as "3 metres per second, every second". It tells you how many $\mathrm{m\,s^{-1}}$ are added to the velocity each second.

Acceleration at Kigali International Airport

A passenger jet taking off from Kigali must reach its take-off speed before the end of the runway. Suppose it reaches $75\,\mathrm{m\,s^{-1}}$ (270 km per hour) from rest in 30 s. Its acceleration is $a = \frac{75 - 0}{30} = 2.5\,\mathrm{m\,s^{-2}}$ : you feel it as the push of the seat on your back. Kigali is about 1500 m above sea level, where the air is thinner, so aircraft need a higher ground speed to take off than at the coast, and so a longer run.

Deceleration matters just as much. When a moto brakes hard, the passenger keeps moving forward; that is why helmets and holding on are compulsory.

Acceleration (Freesciencelessons). Watch the worked calculations using a = (v − u) ÷ t, and note the unit metres per second squared.

✦

Practice

A car speeds up from $5\,\mathrm{m\,s^{-1}}$ to $25\,\mathrm{m\,s^{-1}}$ in 8.0 s. Find its acceleration.
A bus slows from $20\,\mathrm{m\,s^{-1}}$ to $8\,\mathrm{m\,s^{-1}}$ in 4.0 s. Find its acceleration and say what the sign means.
A cyclist starts from rest and accelerates at $0.80\,\mathrm{m\,s^{-2}}$ for 10 s. What is her final velocity?
A moto accelerates from $36\,\mathrm{km\,h^{-1}}$ to $72\,\mathrm{km\,h^{-1}}$ in 5.0 s. Find its acceleration in $\mathrm{m\,s^{-2}}$ .
How long does a train take to reach $30\,\mathrm{m\,s^{-1}}$ from rest if its acceleration is $1.5\,\mathrm{m\,s^{-2}}$ ?
A car moves at a steady $90\,\mathrm{km\,h^{-1}}$ along a straight road. What is its acceleration? Explain.
In a repeat of the ramp experiment, the distances covered in four successive seconds are 0.12 m, 0.36 m, 0.60 m and 0.84 m. Is the acceleration uniform? Find it.

Show answersHide answers

$a = \frac{25 - 5}{8.0} = 2.5\,\mathrm{m\,s^{-2}}$
$a = \frac{8 - 20}{4.0} = -3.0\,\mathrm{m\,s^{-2}}$ . The minus sign means the bus is slowing down: a deceleration of $3.0\,\mathrm{m\,s^{-2}}$ .
$v = 0 + 0.80 \times 10 = 8.0\,\mathrm{m\,s^{-1}}$
$u = 36 \div 3.6 = 10\,\mathrm{m\,s^{-1}}$ , $v = 72 \div 3.6 = 20\,\mathrm{m\,s^{-1}}$ ; $a = \frac{20 - 10}{5.0} = 2.0\,\mathrm{m\,s^{-2}}$
$t = \frac{v - u}{a} = \frac{30 - 0}{1.5} = 20\,\mathrm{s}$
Zero. Its velocity (speed and direction) does not change, so there is no acceleration, however fast it is going.
Yes: each second's distance is 0.24 m more than the one before, so the average speed rises by $0.24\,\mathrm{m\,s^{-1}}$ every second. $a = 0.24\,\mathrm{m\,s^{-2}}$ .

Key points

Acceleration is the rate of change of velocity: $a = \frac{v - u}{t}$ , in $\mathrm{m\,s^{-2}}$ .
Rearranged: $v = u + a t$ .
A negative acceleration means slowing down (deceleration). Constant velocity means zero acceleration.
Uniform acceleration: the velocity changes by the same amount every second.

Sources and credits

2.4: Acceleration, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: ASC Leiden - Rwanda 2021 - 129 - Traffic on a busy road with a girl on the back of a motorcycle - Kigali.jpg by Elvire Eijkman (ASC Leiden), Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: GCSE Physics Revision "Acceleration", Freesciencelessons, embedded from YouTube.

The worked examples, diagrams, Rwandan applications and practice exercises are written by Ganzaa.

1Leaving the traffic lights

2Experiment: a ball rolling down a ramp

3What acceleration means

4Acceleration at Kigali International Airport

Leaving the traffic lights

Experiment: a ball rolling down a ramp

What acceleration means

Acceleration at Kigali International Airport