Question

A car that weighs 1.4 × 10^{4} 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

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