Question

4. Consider a car moving along a straight horizontal road with a
speed of 20ms. If a cofficient of static friction b/w the tyres
and road is 0.5, the shortest distance in which the car can be
stopped is (take g =10ms-2)
(a) 30m
(b) 40m
(c) 72m
(d) 20m​

Answer 1

μ = 0.5

u = 20 m/s

g = 10 m/s²

Draw the free body diagram of the body, as attached.

f = μN

f = μm'g

f = (0.5)(10)m' {Let mass of car be m' kg}

f = 5m'

Net force on car = 0 - 5m' = -5m'

» a = F/m

» a = -5 m/s²

By third equation of motion:

v² = u² + 2as

» 0 = 20² - 2(5)s

» 10s = 400

» s = 40 m

Option (b) ✓