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

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.

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Kiệt Gia 4 years 2021-08-02T05:40:04+00:00 1 Answers 18 views 0

Answers ( )

  1. giabao
    0
    2021-08-02T05:41:29+00:00
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    Answer:

    the answer is below

    Explanation:

    1)

    r=0.84/2=0.42

    The initial velocity (u) = 92 km/h = 92 × 1000 / 3600 m/s = 25.56 m/s.

    the initial angular velocity (\omega_o) = \frac{u}{r}=\frac{25.56}{0.42}=60.84 \ rad/s

    The final velocity (v) = 60 km/h = 60 × 1000 / 3600 m/s = 16.67 m/s.

    the final angular velocity (\omega) = \frac{v}{r}=\frac{16.67}{0.42}=39.68 \ rad/s

    Using:

    w^2=w^2_o+2\alpha \theta\\But\ \theta=77\ rev*2\pi=483.81\ rad\\\\w^2=w^2_o+2\alpha \theta\\\\39.68^2=60.84^2+2\alpha*483.81\\\\1574.5=3701.5+967.62\alpha\\\\967.62\alpha=-2127\\\\\alpha=-2.2 \ rad/s

    2)

    Using:

    w=w_o+\alpha t\\\\39.68=60.84+(-2.)t\\\\39.68-60.84=-2.2t\\\\2.2t=21.16\\\\t=9.62\ s

    3)

    w^2=w^2_o+2\alpha \theta\\But\ w=0\\\\0=w^2_o+2\alpha \theta\\\\0=60.84^2+2(-2.2)\theta\\\\4.4\theta=3701.5\\\\\theta=841.25\ rad=841.25*r \ m=353\ m

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