Question

Chapter 5 Q:6

A car that weighs 1.2 x 104 N is initially moving at a speed of 37 km/h when the brakes are applied and the car is brought to a stop in 20 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 (b) Number Units (c) Number Units (d) Number Units

Answer 1

W= 1.2x10 N m/s u=37 km/h (initial speed) 2om V = 0 ( final speed) - 37x5 18 ( clistence travelled) S- a) Lei = utaas (37**)* + 20 (20) a as because the car is de acceler elevating =) (37x5)-2101(20) (as-2.64m/s) lal=2.64 m/ see? mass of car 1.2 X 10 .xlot g = 1-2xiol 9.8 : : . Force on car ma 1.2x104 x 2.64 9.8 3233.654 N (6) va ut at. 37x5 2.64 t -) t = 3.893 see 18 +

(C) new old initial speed u, initial speed qurah old accelration = new acceleration a (as force doesn't change) eld stopping distance new Stoppind distance = s' dd stopping time=t new stopping time = t' - S v²= u² + zas 2 ų o ₂2lals to ㅋ gals = =ui o (24,1²-alals' 21913'=uur from 0 & ② s = 45

So distance u times times of the before.one. at. ut ( d ? u, lalt =) 3 = U, alt zu -lalt' zul la! t' from o & t' at 2. fucctor of 2 time increases by