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

The angular acceleration of the tires is -2.2 rad/s².

If the car continues to decelerate at this rate, the time required to stop is 27.66 s.

The total distance traveled by the car before stopping is 210.96 revolutions.

The given parameters;

• number of revolutions of the tire, N = 77 rev
• initial linear speed of the car, u  = 92 km/h = 25.56 m/s
• final linear speed of the tire, v = 60 km/h = 16.67 m/s
• diameter of the tire, d = 0.84 m
• radius of the tire, r = 0.42 m

The angular acceleration of the tire is calculated as follows;

$\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\(\frac{16.67}{0.42} )^2 = (\frac{25.56}{0.42} )^2 + 2( 77 \ rev \times \frac{2 \pi \ rad}{1 \ rev} ) \alpha \\\\1575.33 = 3703.59 \ + \ 967.736 \alpha \\\\-2128.26 = 967.736 \alpha\\\\\alpha = \frac{-2128.26}{967.736} \\\\\alpha = - 2.2 \ rad/s^2$

When the car stops, the final angular speed = 0. The time for the motion is calculated as;

$\omega _f = \omega _i + \alpha t\\\\0 = \omega _i + \alpha t\\\\0 = 60.86 + (-2.2)t\\\\0 = 60.86 - 2.2t\\\\2.2t = 60.86\\\\t = \frac{60.86}{2.2} \\\\t = 27.66 \ s$

The total distance traveled by the car before stopping;

$\theta = \omega_i t + \frac{1}{2} \alpha t^2\\\\\theta = (60.86 \times 27.66) \ + \ (0.5 \times -2.2\times 27.66^2)\\\\\theta = 841.8 \ rad\\\\\theta = 841.8 \ rad \times\frac{1 \ rev}{2\pi \ rad} = 133.96 \ rev$

total distance = 133.96 + 77 = 210.96 revolutions.

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