Question

A vehicle moving at 12 m/s stops in 15 m with constant deceleration. Assuming \( g = 10\ \text{m/s}^2 \), compute the coefficient of kinetic friction between tyres and road. (round off to 2 decimal places)

Hint:

Braking deceleration on a level road is directly proportional to the friction coefficient.

Answer 1

Correct Answer: 0.45

Stopping distance relation:
\[ v^2 = u^2 - 2as \] \[ 0 = 12^2 - 2a(15) \] \[ a = \frac{144}{30} = 4.8\ \text{m/s}^2 \] Deceleration is due to kinetic friction:
\[ a = \mu_k g \] \[ \mu_k = \frac{4.8}{10} = 0.48 \] \[ \boxed{0.48} \]