Answered: car make 94 revolutions as the car…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

The tires of a car make 94 revolutions as the car reduces its speed uniformly from 85.0 km/h to 65.0 km/h. The tires have a diameter of 0.84 m.

Part A

What was the angular acceleration of the tires?

Express your answer using two significant figures.

α = ____ rad/s^2

Part B

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

Express your answer to two significant figures and include the appropriate units.

t = ________

Part C

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

Express your answer to three significant figures and include the appropriate units.

s =________

Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .

Expert Solution

Step 1: solving part A

To calculate the angular acceleration of the tires, we can use the following equation:

α = (ω_f - ω_i) / Δt

where:

α is the angular acceleration in rad/s^2
ω_i is the initial angular velocity in rad/s
ω_f is the final angular velocity in rad/s
Δt is the time interval in s

First, we need to convert the speeds to m/s:

v_i = 85.0 km/h * 5/18 = 23.6 m/s
v_f = 65.0 km/h * 5/18 = 18.1 m/s

Next, we need to calculate the angular velocity of the tires at each speed:

ω_i = v_i / r = 23.6 m/s / 0.42 m = 56.2 rad/s
ω_f = v_f / r = 18.1 m/s / 0.42 m = 43.1 rad/s

Finally, we can calculate the angular acceleration:

α = (ω_f - ω_i) / Δt = (43.1 rad/s - 56.2 rad/s) / 94 rev * 2π rad/rev = -0.110 rad/s^2

Therefore, the angular acceleration of the tires is -0.110 rad/s^2.

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