The tires of a car make 95 revolutions as the car reduces its speed uniformly fr

ID: 1395310 • Letter: T

Question

The tires of a car make 95 revolutions as the car reduces its speed uniformly from 90.0 km/h to 65.0 km/h . The tires have a diameter of 0.80 m .

- What was the angular acceleration of the tires?

Express your answer using two significant figures.

- 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.

- How far does the car go? Find the total distance.

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

Please show how to work it out as well, not only the answer.

Here ,

number of revolutions = 95 revv

angle = 95 * 2pi

angle = 596.8 rad

intial speed , u = 90 km/h = 25 m/s

final speed , v = 65 km/h = 18.1 m/s

As v = r*w

initial angular speed , wi = 25/(0.80/2)

wi = 62.5 rad/s

final angular speed , wf = 18.1/(0.80/2)

wf = 45.3 rad/s

Now, let the angular acceleration is alpha ,

using third equation of motion ,

wf^2 - wi^2 = 2 *alpha * angle

45.3^2 - 62.5^2 = 2 * 596.8 * alpha

alpha = -1.55 rad/s^2

the angular acceleration is -1.55 rad/s^2

let the time required to stop is t ,

using first equation of motion

0 = 45.3 - 1.55 * t

t = 29.2 s

the time taken to stop is 29.2 s

total distance = (25^2 - 18.1^2)/(2 * 0.4 * 1.55)

total distance =239.8 m

the total distance travelled is 239.8 m

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