Question

The driver of a bus moving with a velocity of 72 kmph observes a boy walking across the road at a distance of 50 m in front of the bus and decelerates the bus at 5 \( \text{ms}^{-2} \) by applying brakes and is just able to avoid an accident. The reaction time of the driver is

• A. 4 s
• B. 3.5 s
• C. 0.5 s
• D. 4.5 s

Hint:

Remember: Total Stopping Distance = (Velocity \(\times\) Reaction Time) + \( \frac{\text{Velocity}^2}{2 \times \text{Deceleration}} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The total stopping distance is the sum of the distance traveled during the reaction time (reaction distance) and the distance traveled while braking (braking distance).
Total Distance \( S = S_{\text{reaction}} + S_{\text{braking}} \).
Step 2: Key Formula or Approach:
1. Convert velocity: \( v = 72 \times \frac{5}{18} = 20 \, \text{m/s} \).
2. Braking distance \( S_b = \frac{v^2}{2a} \) (from \( v_f^2 - v_i^2 = 2as \)).
3. Reaction distance \( S_r = v \times t_r \).
4. Total distance \( S = 50 \) m.

Step 3: Detailed Explanation:
Initial velocity \( u = 20 \, \text{m/s} \). Deceleration \( a = 5 \, \text{m/s}^2 \). Calculate Braking Distance (\( S_b \)): Using \( 0^2 - u^2 = 2(-a)S_b \): \[ S_b = \frac{u^2}{2a} = \frac{(20)^2}{2 \times 5} = \frac{400}{10} = 40 \, \text{m} \] Given Total Distance \( S = 50 \, \text{m} \). The bus travels this total distance before stopping. \[ S = S_r + S_b \] \[ 50 = S_r + 40 \implies S_r = 10 \, \text{m} \] Calculate Reaction Time (\( t_r \)): Since the bus moves at constant velocity \( u \) during the reaction time: \[ S_r = u \times t_r \] \[ 10 = 20 \times t_r \] \[ t_r = \frac{10}{20} = 0.5 \, \text{s} \]
Step 4: Final Answer:
The reaction time is 0.5 s.