Question

A car, starting from rest, accelerates at the rate f through a distance , then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 15 , then

(a) * = 1/6ft2

(b) * = ft

(c) * = 1/4ft2

(d) * = 1/72ft2

Answer 1

*Correct option (d) = 1/72ft2**

Explanation:

Distance from A to B = * = 1/2ft21

Distance from B to C = (ft1)t

Distance from C to D = u2/2a = (ft1)2/2(f/2)

= ft12 = 2S

⇒ * + ft1t + 2S = 15S

⇒ ft1t = 12S ....(i)

1/2ft21 = * .....(ii)

Dividing (i) by (ii), we get t1 = t/6

⇒ * = 1/2f(t/6)2 = ft2/72