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?
Yes, the bus will be able to stop before reaching the obstacle.
Explanation
The total distance the bus travels before stopping is the sum of the distance covered during the driver's reaction time and the distance covered after applying the brakes. The bus travels at a constant velocity during the reaction time, covering 5 m. Once the brakes are applied, it covers another 20 m before coming to a halt. The total stopping distance is 25 m, which is less than the 30 m distance to the obstacle, meaning the bus will stop safely.
Solution Steps
Step 1: Convert initial velocity to m/s: u = 36 km h⁻¹ = 36 × (1000 m / 3600 s) = 10 m s⁻¹
Step 2: Calculate distance covered during reaction time: s₁ = u × t = 10 m s⁻¹ × 0.5 s = 5 m
Step 3: Calculate braking distance using equation of motion v² - u² = 2as: 0² - (10)² = 2(-2.5)s, which gives -100 = -5s, so s₂ = 20 m
Step 4: Calculate total stopping distance: Total distance = s₁ + s₂ = 5 m + 20 m = 25 m
Step 5: Compare total distance with obstacle distance: 25 m < 30 m, so the bus will stop before the obstacle.