A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle?

Initial velocity of the bus,

$$u = 36 \text{ km h}^{-1}$$

$$u = \frac{36 \times 1000}{3600} = 10 \text{ ms}^{-1}$$

Distance of obstacle from the bus = 30 m

Reaction time of driver, $t = 0.5 \text{ s}$

Retardation after applying brake, $a = -2.5 \text{ ms}^{-2}$

During the reaction time, the driver does not apply brakes, so the bus continues to move with the same velocity.

Distance travelled during reaction time,

$$s_1 = vt$$

$$s_1 = 10 \times 0.5$$

$$s_1 = 5 \text{ m}$$

Now, distance left before reaching the obstacle,

$$30 - 5 = 25 \text{ m}$$

Then, stopping distance after brakes are applied.

$$v^2 = u^2 + 2as$$

Since the bus stops,

$$v = 0$$

$$0 = (10)^2 + 2(-2.5)s$$

$$0 = 100 - 5s$$

$$5s = 100$$

$$s = 20 \text{ m}$$

Thus, the bus requires 20 m to stop after applying brakes.

Total distance travelled before stopping,

$$= 5 + 20$$

$$= 25 \text{ m}$$

Since the obstacle is 30 m ahead, the bus stops 5 m before the obstacle. Therefore, the bus will be able to stop before reaching the obstacle.