Question

A car is moving along a straight horizontal road with speed of \(20\,\mathrm{m/s}\). If the coefficient of friction between tyres and roads is \(0.4\), then the shortest distance in which the car can stop is (Take \(g=10\,\mathrm{m/s^2}\)).

• A. \(60\,\mathrm{m}\)
• B. \(40\,\mathrm{m}\)
• C. \(80\,\mathrm{m}\)
• D. \(50\,\mathrm{m}\)

Hint:

For maximum braking on a horizontal road, \[ a=\mu g, \] and the stopping distance is \[ \boxed{s=\frac{u^2}{2\mu g}}. \] This formula is commonly used in Engineering Mechanics and Transportation Engineering.

Answer 1

The Correct Option is D

Concept: The shortest stopping distance occurs when the braking force is equal to the maximum frictional force. The maximum deceleration is \[ a=\mu g, \] where \[ \mu=\text{coefficient of friction},\qquad g=\text{acceleration due to gravity}. \] The stopping distance is obtained from the equation of motion \[ v^2=u^2+2as. \] Since the car comes to rest, \[ v=0. \]

Step 1: Calculate the maximum deceleration. Given, \[ u=20\,\mathrm{m/s}, \] \[ \mu=0.4, \] \[ g=10\,\mathrm{m/s^2}. \] Therefore, \[ a=-\mu g =-(0.4)(10) =-4\,\mathrm{m/s^2}. \] (The negative sign indicates retardation.)

Step 2: Apply the equation of motion. Using \[ v^2=u^2+2as, \] \[ 0=20^2+2(-4)s. \] Hence, \[ 400=8s, \] which gives \[ s=\frac{400}{8}=50\,\mathrm{m}. \] Therefore, \[ \boxed{s=50\,\mathrm{m}.} \] Thus, the correct option is \[ \boxed{(D)\ 50\,\mathrm{m}.} \]