Question

The braking efficiency for a vehicle moving with a speed of 18 kmph, having a lag distance of 14 m and coefficient of longitudinal friction of 0.36 is .......

• A. 25.28%
• B. 25.4%
• C. 25.6%
• D. 25.8%

Hint:

Braking efficiency drops when lag distance increases—optimize driver reaction time and tire-road friction to maximize safety.

Answer 1

The Correct Option is A

Step 1: Write down the known values
Speed, $V = 18$ kmph = $18 \times \frac{1000}{3600} = 5$ m/s
Lag distance, $L = 14$ m
Coefficient of friction, $f = 0.36$
Step 2: Calculate total stopping distance using the general formula
\[ \text{Stopping distance} = \text{Lag distance} + \text{Braking distance} \] \[ \text{Braking distance} = \frac{V^2}{2gf} = \frac{(5)^2}{2 \times 9.81 \times 0.36} = \frac{25}{7.0632} \approx 3.54 \text{ m} \] \[ \text{Total stopping distance} = 14 + 3.54 = 17.54 \text{ m} \] Step 3: Calculate ideal braking distance (if there were no lag)
\[ \text{Ideal stopping distance} = \frac{V^2}{2gf} = 3.54 \text{ m} \] Step 4: Compute braking efficiency using the formula:
\[ \text{Braking Efficiency (BE)} = \left( \frac{\text{Ideal stopping distance}}{\text{Actual stopping distance}} \right) \times 100 = \left( \frac{3.54}{17.54} \right) \times 100 \approx 25.28% \]