Question

The tires of a car make 56 revolutions as the car reduces its
speed uniformly from 93 km/h to 50 km/h . The tires have a diameter
of 0.76 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?

Answer #1

(A) = 56 revolutions = 56*2*pi = 351.8 radians

v1 = 93 km/h = (93 * 1000) / (60*60) = 25.83 m/s

So, w1 = v1 / r = 25.83/0.38 = 68 rad/s

v2 = 50 km/h = (50*1000) / (60*60) = 13.89 m/s

=> w2 = 13.89/0.38 = 36.5 rad/s

Use the expression -

w2^2 = w1^1 + 2**

=> 36.5^2 = 68^2 + 2**351.8

=> = (36.5^2 - 68^2) / (2*351.8) = -4.68 rad/s^2

So, the magnitude of the angular acceleration = 4.68 rad/s^2

(B) When the car stop, w2 = 0

use the expression -

w2 = w1 + *t

=> 0 = 68 + (-4.68)*t

=> t = 68 / 4.68 = 14.5 sec.

So time of stopping of the car = 14.5 s.

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speed uniformly from 120 km/h to 50 km/h. The tires have a diameter
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(b) If the car continues to decelerate at this rate, how much more
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What was the angular acceleration of the tires?
If the car continues to decelerate at this rate, how much more
time is required for it to stop?
If the car continues to decelerate at this rate, how far does it
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B) If the car continues to decelerate at this rate, how much
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a= ? rad/s^2
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time is required for it to stop?
If the car continues to decelerate at this rate, how far does it
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Part A What was the angular acceleration of the
tires?
Part B If the car continues to decelerate at
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Part C If the car continues to decelerate at
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more time is required for it to stop? C) How far does the car go?
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