Problem 2 : During a very quick stop, a car decelerates at 6.4 m/s 2 . Assume th

ID: 2240531 • Letter: P

Question

Problem 2: During a very quick stop, a car decelerates at 6.4 m/s2. Assume the car is moving in the positive horizontal direction.

Randomized Variablesat = 6.4
r = 0.26 m
?0 = 92 rad/s
Part (a) What is the angular acceleration of its 0.26 m-radius tires, assuming they do not slip on the pavement in rad/s2?
Part (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 92 rad/s ?
Part (c) How long does the car take to stop completely in seconds?
Part (d) What distance does the car travel in this time in meters?
Part (e) What was the car During a very quick stop, a car decelerates at 6.4 m/s2. Assume the car is moving in the positive horizontal direction.

a). Angular acceleration=(r *at)

=0.26*6.4

= 1.664 rad/s^2

b). we know W1^2-W0^2=2*alph*theta

where W1-Final angular velocity

W0-Initial angular velocity

alpha-angular acceleration

theta-andular displacement

therefor theta=(92^2/(2*1.644))

=2574.2 rad

number of revolutions=(2574.2/(2*pi))

=409.68 revolutions

c). we know W1=W0+alpha*t

but final velocity=0

t=(W0/alpha)

=55.9 sec or 56 sec approx

d).Car's initial velocity in meters=(W0/r)

=92/0.26

= 353.84 m/s

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