Question

A car that weighs 1.4 × 10⁴ N is initially moving at a speed of 35 km/h when the brakes are applied and the car is brought to a stop in 18 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

Answer 1

m= W/g = (1.4 x 10^4)/9.8

m = 0.143 x 10^4 kg

vi = 35 x 1000 m / 3600s = 9.72 m/s

Applying work - energy theorem,

Work done = change in KE

- F.d = m vf^2 /2 - m vi^2 / 2

- 18 F = 0 - (0.143 x 10^4)(9.72^2)/2

F = 3755 N .....Ans

(B) a = F/m = 2.63 m/s^2

vf = vi + a t

0 = 9.72 - 2.63 t

t = 3.7 s

(C) same force means same acceleration.

vf^2 - vi^2 = 2 a d

0 - v^2 =2(-a)(d)

v^2 = 2 a d

v -> doubled then d -> 4 times

d = 18 x 4 = 72 m

(D) vf = vi + a t

0 = v - a t

t = v / a

v -> doubled then t->doubled

t = 3.7 x 2 = 5.4 sec