A car that weighs 1.3 × 10 4 N is initially moving at a speed of 42 km/h when th

Question

A car that weighs 1.3 × 104 N is initially moving at a speed of 42 km/h when the brakes are applied and the car is brought to a stop in 15 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.)

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A car that weighs 1.3 x 104 Nis initially moving at a speed of 42 km/h when the brakes are applied and the car is brought to a stop in 15 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.) (a) Number Units Units s (b) Numbe 2.57 (c) Numbe UnitsEl (d) Number

Explanation / Answer

a)

v^2 = u^2 + 2as

u = 42 km/h = 42/3.6 = 11.67 m/s

v = 0

s = 15 m

So, 0^2 = 11.67^2 - 2*(a)*15

So, a = 4.54 m/s2

So, magnitude of force = m*a = (1.3*10^4/9.8)*4.54 = 6022.4 N<---------answer

b)

Using the equation of motion,

v = u + at

So, 0 = 11.67 - 4.54*t

So, t = 2.57 s <--------answer

c)

v^2 = u^2 - 2as

So, 0^2 = u^2 - 2as

So, s = u^2/2a

So, as u becomes 2 times , s becomes 2^2 = 4 times <----answer

d)

v = u + at

So, 0 = u - a*t

So, t = u/a

As u changes 2 times, t also changes 2 times <-------answer

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