Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s.

Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).

For car A:

$$u = 0, v = 5 \text{ ms}^{-1}, t = 5 \text{ s}$$

Acceleration of car A,

$$\begin{aligned} a &= \frac{v-u}{t} \\ a &= \frac{5-0}{5} \\ a &= 1 \text{ ms}^{-2} \end{aligned}$$

For car B:

$$u = 0, v = 3 \text{ ms}^{-1}, t = 10 \text{ s}$$

Acceleration of car B,

$$\begin{aligned} a &= \frac{3-0}{10} \\ a &= 0.3 \text{ ms}^{-2} \end{aligned}$$

To plot the velocity-time graph:

(1) Draw time on the X-axis and velocity on the Y-axis.

(2) For car A, join the points (0, 0) and (5, 5).

(3) For car B, join the points (0, 0) and (10, 3).

Therefore, the displacement of car A is 12.5 m and the displacement of car B is 15 m.

The displacement is equal to the area under the velocity-time graph.

For car A:

$$\begin{aligned} \text{Displacement} &= \frac{1}{2} \times 5 \times 5 \\ &= 12.5 \text{ m} \end{aligned}$$

For car B:

$$\begin{aligned} \text{Displacement} &= \frac{1}{2} \times 10 \times 3 \\ &= 15 \text{ m} \end{aligned}$$