Question

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?

Answer 1

Given,

Initial velocity (u) = 36 km h^–1

Convert km h^–1 to m s^–1: multiply by `5/18`

u = 36 × `5/18 = 10` m s^–1

Step 1: Distance travelled during the reaction time (before brakes are applied)

The bus travels at a steady speed of 10 m s^–1 during the 0.5 s reaction time.

Reaction distance = velocity × time

= 10 × 0.5

= 5 m

Step 2: Distance travelled after the brakes are applied

u = 10 m s^–1, v = 0, a = −2.5 m s^–2

Using the third kinematic equation,

v² = u² + 2as

(0)² = (10)² + 2 × (−2.5) × s

0 = 100 – 5s

⇒ s = `100/5 = 20` m

∴ Total distance travelled before stopping = reaction distance + braking distance

= 5 + 20

= 25 m

Given that the obstacle’s distance (30 m) is more than the overall stopping distance (25 m),

Yes, the bus can stop five meters short of the obstruction before it reaches it.