Question

A car is travelling at a speed of 60 km/hr on a section of a National Highway having a downward gradient of 2\%. The driver of the car suddenly observes a stopped vehicle on the car path at a distance 130 m ahead, and applies brake. If the brake efficiency is 60\%, coefficient of friction is 0.7, driver’s reaction time is 2.5 s, and acceleration due to gravity is 9.81 m/s\(^2\), the distance (in meters) required by the driver to bring the car to a safe stop lies in the range

• A. 126 to 130
• B. 41 to 45
• C. 33 to 37
• D. 75 to 79

Hint:

To calculate the stopping sight distance, always consider both the reaction distance and braking distance, accounting for road gradient and brake efficiency.

Answer 1

The Correct Option is D

Step 1: Given data: \[ \text{Initial speed } (u) = 60 \text{ km/hr} = \frac{60 \times 1000}{3600} = 16.67 \text{ m/s} \] \[ \text{Reaction time } (t_r) = 2.5 \text{ s} \] \[ \text{Coefficient of friction } (\mu) = 0.7, \quad \text{Brake efficiency} = 60\% \Rightarrow \text{effective } \mu = 0.7 \times 0.6 = 0.42 \] \[ \text{Gradient } (G) = 2\% = 0.02 \] \[ \text{Acceleration due to gravity } (g) = 9.81 \text{ m/s}^2 \] Step 2: Calculating the reaction distance: \[ \text{Reaction distance } = u \times t_r = 16.67 \times 2.5 = 41.675 \text{ m} \] Step 3: Calculating the braking distance using the formula: \[ \text{Braking distance } = \frac{u^2}{2 g (\mu \pm G)} \] Substituting the values: \[ = \frac{(16.67)^2}{2 \times 9.81 \times (0.42 + 0.02)} \] \[ = \frac{278.08}{2 \times 9.81 \times 0.44} \] \[ = \frac{278.08}{8.6352} \approx 32.2 \text{ m} \] Step 4: Total stopping distance: \[ \text{Total stopping distance} = \text{Reaction distance} + \text{Braking distance} \] \[ = 41.675 + 32.2 = 73.875 \approx 75 \text{ m} \] Conclusion: The total stopping distance lies in the range 75 to 79, which corresponds to option (D).