A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ as shown in the figure for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

Initial velocity,

$$\begin{aligned} u &= 54 \text{ km h}^{-1} \\ u &= \frac{54 \times 1000}{3600} = 15 \text{ ms}^{-1} \end{aligned}$$

Final velocity,

$$\begin{aligned} v &= 36 \text{ km h}^{-1} \\ v &= \frac{36 \times 1000}{3600} = 10 \text{ ms}^{-1} \end{aligned}$$

Time taken,

$$t = 36 \text{ s}$$

Using the equation,

$$\begin{aligned} s &= \frac{u+v}{2} \times t \\ s &= \frac{15+10}{2} \times 36 \\ s &= \frac{25}{2} \times 36 \\ s &= 12.5 \times 36 \\ s &= 450 \text{ m} \end{aligned}$$

Therefore, the distance travelled by the truck during this time is 450 m.