Question

A car, starting from rest, accelerates at the rate f through a distance S, 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 S , then

(a) S = 1/6ft²

(b) S = ft

(c) S = 1/4ft²

(d) S = 1/72ft²

Answer 1

Correct option (d) S = 1/72ft²

Explanation:

Distance from A to B = S = 1/2ft²₁

Distance from B to C = (ft₁)t

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

= ft₁² = 2S

⇒ S + ft₁t + 2S = 15S

⇒ ft₁t = 12S ....(i)

1/2ft²₁ = S .....(ii)

Dividing (i) by (ii), we get t₁ = t/6

⇒ S = 1/2f(t/6)² = ft²/72