Question

The tires of a car make 93 revolutions as the car reduces its speed uniformly from 92.0 km/h to 60.0 km/h. The tires have a diameter of 0.86 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?

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

Answer 1

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Answer 2

To solve this problem, we'll use the following formulas:

1. Angular acceleration (α) = (ωf - ωi) / t

2. Final angular velocity (ωf) = (2π * Nf) / t

3. Initial angular velocity (ωi) = (2π * Ni) / t

4. Time (t) = (Nf - Ni) / ωf

5. Distance (s) = (ωi * t) + (0.5 * α * t^2)

Given:

• Number of revolutions (Nf) = 93

• Initial speed (ωi) = 92.0 km/h

• Final speed (ωf) = 60.0 km/h

• Diameter of tires (d) = 0.86 m

First, we need to convert the speeds to radians per second: ωi = (92.0 km/h) * (1000 m/km) / (3600 s/h) = 25.6 m/s ωf = (60.0 km/h) * (1000 m/km) / (3600 s/h) = 16.7 m/s

A) Angular acceleration (α): α = (ωf - ωi) / t

To find the time (t), we can use the formula: t = (Nf - Ni) / ωf

Substituting the values: t = (93 - 0) / (16.7 m/s) = 5.57 s

Now we can calculate α: α = (ωf - ωi) / t = (16.7 m/s - 25.6 m/s) / 5.57 s = -1.60 rad/s^2 (negative sign indicates deceleration)

B) Time to stop: The time required for the car to stop is twice the time it took to reduce the speed from 92.0 km/h to 60.0 km/h. Therefore, the total time to stop is 2 * 5.57 s = 11.14 s.

C) Distance traveled: To find the total distance traveled, we'll use the formula: s = (ωi * t) + (0.5 * α * t^2)

Substituting the values: s = (25.6 m/s * 5.57 s) + (0.5 * -1.60 rad/s^2 * (5.57 s)^2) = 141.7 m

Therefore, the car travels a total distance of approximately 141.7 meters.