A car that weighs 1.0 × 10 4 N is initially moving at a speed of 36 km/h when th

Question

A car that weighs 1.0 × 104 N is initially moving at a speed of 36 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.)

Explanation / Answer

Here ,

mass of car , m = W/g

m = 1 *10^4/9.8

m = 1020.5 Kg

initial speed , u = 36 km/hr = 10 m/s

a) let the force is F

Using work energy theorum

F * d = - 0.5 m * v^2

F * 18 = 0.5 * 1020.5 * 10^2

F = 2834 N

the magnitude of force is 2834 N

b)

let the time taken is t

acceleration ,a = F/m

a = 2834/1020.5

a = 2.78 m/s^2

Using first equation of motion

v = u + a * t

0 = 10 - 2.78* t

t = 3.6 s

the time taken is 3.6 s

c)

For the same acceleration , as the stopping distance is

d = u^2/(2 * a)

for double speed ,

dnew = 4 * d

dnew = 4 * 18 = 72 m

new distance will be 4 times

d)

t = u/a

for double initial speed ,

tnew = 2 * t

new time will be 2 times

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