Question

The tires of a car make 77 revolutions as the car reduces its speed uniformly from 92.0 km/h to 60.0 km/h. The tires have a diameter of 0.84 m.

1. What was the angular acceleration of the tires?

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

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

Answer 1

Part A.

Given that Vi = 92.0 km/hr = 25.56 m/sec

Vf = 60.0 km/hr = 16.67 m/sec

D = diameter of tire = 0.84 m

R = radius of tire = 0.42 m

Since V = w*R, So

Initial angular velocity = wi = Vi/R = 25.56/0.42 = 60.86 rad/sec

final angular velocity = wf = Vf/R = 16.67/0.42 = 39.69 rad/sec

theta = 77 rev = 77*2*pi = 438.805 rad

Now Using 3rd rotational kinematic equation:

wf^2 = wi^2 + 2*alpha*theta

alpha = angular acceleration = (wf^2 - wi^2)/(2*theta)

alpha = (39.69^2 - 60.86^2)/(2*438.805)

alpha = -2.4255 = -2.43 rad/sec^2

Part B.

when wf = 0, then

Now wi = 39.69 rad/sec

Using 1st kinematic equation:

wf = wi + alpha*t

t = (wf - wi)/alpha

t = (0 - 39.69)/(-2.4255)

t = 16.36 sec

Part C.

When car makes 77 rev, then

d1 = R*theta = 0.42*77*2*pi = 203.2 m

Using 2nd kinematic equation for distance traveled in next 16.36 sec:

theta1 = wi*t + (1/2)*alpha*t^2

theta1 = 39.69*16.36 + (1/2)*(-2.4255)*16.36^2 = 324.74 rad

d2 = R*theta1 = 0.42*324.74 = 136.4 m

So total stopping distance = 203.2 + 136.4 = 339.6 m

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