Question

The tires of a car make 77 revolutions as the car reduces its speed uniformly from 87.0 km/h to 55.0 km/h. The tires have a diameter of 0.90 m.

A) What was the angular acceleration of the tires?

B) If the car continues to decelerate at this rate, how much more time is required for it to stop?

C) If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Answer #1

a) we have v_{0}= 87 km/h= (87/3.6)m/s=24.2m/s

v_{1}=55km/h=(55/3.6)m/s=15.3m/s

_{0}=v_{0}/r...,
r=d/2=.45m

=24.2/.45=53.8 rad/sec

=v/r

=15.3/.45=34 rad/s

by applying the formula

^{2}=_{0}^{2}+2

where= angular acceleration

=
^{2}-_{0}^{2}/2

=34^{2}-53.8^{2}/2* 77*2pi

=-1.78rad/s^{2} ( dont forget negative sign)

so a) -1.78 rad/s^{2}

b) we have to apply the formula

v=u+at

v=0

u+at=0

34+(-1.78t)=0

t=34/1.78=19.1 sec

b)19.1sec

c) d=77*0.90*pi=217.60m

t=2d/v_{0}+v)

=2*217.60/24.2+15.3

=11.01 sec

slowing distance =217.6 m

and the total stopping distance =1/2v(11.01+19.10)

=1/2*24.2*30.11

=364.33m

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