Equations of motion and velocity–time graphs · Kinematics & dynamics

A body moving in a straight line with constant acceleration is described by five quantities: the starting velocity $u$ , the final velocity $v$ , the acceleration $a$ , the time $t$ and the displacement $s$ . A negative acceleration slows the body down: it is called a retardation (or deceleration).

The equations of motion

On a velocity–time graph, constant acceleration is a straight line. Its gradient is $a$ and the area under it is $s$ :

Constant accelerationgradient = a = (v − u) ÷ t; area = s = ½(u + v)t

Those two facts give the four equations. Pick the one that leaves out the quantity you neither know nor want:

\begin{aligned} v &= u + at & s &= \tfrac12(u + v)t \\ s &= ut + \tfrac12at^2 & v^2 &= u^2 + 2as \end{aligned}

Keep the units the same throughout: convert km/h to m/s (divide by 3.6) or minutes to hours before you start.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q7

A car travelling at $15\text{ m s}^{-1}$ is brought to rest with uniform retardation in 5 seconds. Find the:

retardation;

distance travelled before coming to rest.

List what you know
- ${u = 15}$ , ${v = 0}$ (at rest) and ${t = 5}$ .
Think first. u, v and t are known. Which equation finds a?
The retardation
- ${v = u + at}$ : ${0 = 15 + 5a}$ , so ${a = -3}$ m/s².
- The negative sign means slowing down: the retardation is ${3}$ m/s².
The distance
- ${s = \frac12(u + v)t = \frac12(15 + 0)(5) = 37.5}$ m.
Think first. Use the equation without a.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q8 (a)

A car moving on a straight road with constant acceleration has a velocity of $20\text{ km h}^{-1}$ at an instant. If 15 minutes later it had a velocity of $50\text{ km h}^{-1}$ , find the acceleration of the car.

Match the units
- ${15\text{ min} = \frac14\text{ h}}$ .
Think first. The speeds are in km/h. What is 15 minutes in hours?
The acceleration
- ${v = u + at}$ : ${50 = 20 + \frac14a}$ .
- So ${a = 4 \times 30 = 120}$ km/h².

More: the equations of motion

Journeys on a velocity–time graph

When a journey has stages (speeding up, a steady speed, slowing down), draw the velocity–time graph. The distance is the whole area under it, and the average speed is the total distance divided by the total time.

A journey in three stagesdistance = area; average speed = distance ÷ time

Distance from a velocity–time graphSet the times and the top speed

192 mdistance (area)10.67 m/saverage speed

speed-up time (s) = 4top speed (m/s) = 16steady time (s) = 6slow-down time (s) = 8

Area = ½ × 4 × 16 + 6 × 16 + ½ × 8 × 16 = 32 + 96 + 64 = 192 m. Average speed = 192 ÷ 18 = 10.67 m/s. Acceleration = 16 ÷ 4 = 4 m/s²; retardation = 16 ÷ 8 = 2 m/s².

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q14

A body moving at $20\text{ m s}^{-1}$ accelerates uniformly at $2\frac12\text{ m s}^{-2}$ for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest $t$ seconds after with uniform retardation. The ratio of the acceleration to the retardation is $3 : 4$ . Sketch the velocity–time graph of the motion.

Find the value of $t$ .

Find the total distance of the journey.

The sketch
- After 4 s: ${v = 20 + 2.5 \times 4 = 30}$ m/s.
- The graph starts at 20 (not at 0), rises to 30 at ${t = 4}$ , is level until ${t = 12}$ , then falls to 0.
Think first. The body is already moving at 20 m/s. Where does the graph start?
The retardation and t
- ${2.5 : r = 3 : 4}$ , so ${r = \frac{4 \times 2.5}{3} = \frac{10}{3}}$ m/s².
- ${t = 30 \div \frac{10}{3} = 9}$ s.
Think first. Acceleration : retardation = 3 : 4.
The distance: the area
- Trapezium: ${\frac12(20 + 30)(4) = 100}$ .
- Rectangle: ${30 \times 8 = 240}$ . Triangle: ${\frac12(30)(9) = 135}$ .
- Total: ${100 + 240 + 135 = 475}$ m.

More: velocity–time graphs

NECO 2023 · Paper 2 · Q13A car starts from town $A$ and accelerates uniformly for 4 minutes until it reaches a speed of $35\text{ ms}^{-1}$ , which …

Your turn

WAEC 2020 · Paper 2 · Q15 (a)

(a)
A particle starts from rest at a point $O$ and accelerates uniformly for 3 minutes until it attains a velocity of $133\text{ m s}^{-1}$ . It moves with this velocity for 3 minutes and then retards uniformly for another 2 minutes before coming to rest at $P$ . (i) Sketch the velocity–time graph for the motion. (ii) Calculate the: (α) total distance covered by the particle; (β) average velocity of the particle.
Separate values with commas, e.g. 3, −2

Try it on a graph

Velocity–time graph: the area under it is the distance.

Worked solution (try it first)

(a)(i)

Times: $3\text{ min} = 180\text{ s}$ and $2\text{ min} = 120\text{ s}$ .
The graph rises from 0 to $133$ over $0$ – $180\text{ s}$ , is level to $360\text{ s}$ , then falls to 0 at $480\text{ s}$ .

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