A car starts from rest and accelerates uniformly to 20 ms⁻¹ in 5 seconds. It then travels at 20 ms⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.

The motion of the car takes place in three stages.

First stage: Car accelerates from rest

Given,

$$\begin{aligned} u &= 0 \text{ ms}^{-1} \\ v &= 20 \text{ ms}^{-1} \\ t &= 5 \text{ s} \end{aligned}$$

Distance travelled during acceleration,

$$\begin{aligned} s_1 &= \frac{(u+v)}{2} \times t \\ s_1 &= \frac{(0+20)}{2} \times 5 \\ s_1 &= 10 \times 5 \\ s_1 &= 50 \text{ m} \end{aligned}$$

Second stage: Car moves with constant velocity

$$\text{Velocity } (v) = 20 \text{ ms}^{-1}$$

$$\text{Time } (t) = 10 \text{ s}$$

Distance travelled,

$$\begin{aligned} s_2 &= vt \\ s_2 &= 20 \times 10 \\ s_2 &= 200 \text{ m} \end{aligned}$$

Third stage: Car slows down and stops

Given,

$$\begin{aligned} u &= 20 \text{ ms}^{-1} \\ v &= 0 \text{ ms}^{-1} \\ t &= 6 \text{ s} \end{aligned}$$

Distance travelled during braking,

$$s_3 = \frac{(u+v)}{2} \times t$$

$$s_3 = \frac{(20+0)}{2} \times 6$$

$$s_3 = 10 \times 6$$

$$s_3 = 60 \text{ m}$$

Therefore,

Total distance travelled = $s_1 + s_2 + s_3$

$$= 50 + 200 + 60$$

$$= 310 \text{ m}$$

Therefore, the total distance travelled by the car is 310 m.