Given data:
W = 20KN, u = 70Kmphr = 19.44m/sec, v = 0, t = ?
Consider FBD of the car as shown in fig,
∑V = 0, R = W ...(i)
∑H = 0, Fr = 0 Fr = µR ...(ii)
Here net force is the frictional force
i.e. F = Fr ma = µR = µmga = µg ...(iii)
(1) on concrete road for which µ = 0.75
a = µg = 0.75 X 9.81 = 7.3575
a = 7.35 m/sec2 ...(iv)
Using the relation v = u – at
0 = 19.44 – 7.35t
t = 2.64 seconds
(1) On ice for which µ = 0.08
a = µg = 0.08 X 9.81 = 0.7848
a = 0.7848m/sec2
Using the relation v = u – at
0 = 19.44 – 0.7848t
t = 24.77 seconds
