A car that weighs 1.5 × 104 N is initially moving at a speed of 43 km/h when the

Question

A car that weighs 1.5 × 104 N is initially moving at a speed of 43 km/h when the brakes are applied and the car is brought to a stop in 19 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.)

Explanation / Answer

Weight, mg = 1.5 x104 N

m = (1.5 x 104)/g

g = 9.8 m/s 2

m = 15.92 kg

initial velocity, u = 43 km/h = (43 x 1000)/3600 m/s

= 11.94 m/s

final velocity, v = 0

stopping distance, d = 19m

acceleration, a = (v 2 - u 2)/2d

a = (0 - 11.942)/(2x19)

a = - 3.75 m/ s 2

a) Negative signs shows that this acceleration is decreasing velocity, since magnitude is asked we will drop this negative sign.

Force, F = ma

F = 15.92 x 3.75

= 59.7 N

b) time, t = (v - u)/a

t = (0 - 11.94)/(-3.75)

t = 3.18 s

c) u = 2 x 11.94 = 23.88 m/s

new distance, S = (v 2 - u 2 )/2a

S = (0 - 23.882)/(2x (-3.75))

S = 76.03 m

Note that doubling the initial speed require four times stopping distance with same braking force.

d) new time, T = (0 - 23.88)/ (-3.75)

T = 6.37 s

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