Question

Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is $[g=10m/s^{-2}]$

• A. 30 m
• B. 40 m
• C. 72 m
• D. 20 m

Answer 1

The correct option is (A) 40 m

Step 1, Given data

Speed = 72 km/h = 20 m/s

Coefficient of the kinetic friction = 0.5

Acceleration = $0.5\times 10=-5$ m/$s^2$

Step 2, Finding the distance

From the equation of motion

We know,

$v^2=u^2+2as$

Putting all the values

$0^2={20}^2-2\times 5\times s$

After solving

s = 40 m

Hence the distance traveled before stopping is 40 m