Problem 8.22 The tires of a car make 86 revolutions as the car reduces its speed

Question

Problem 8.22 The tires of a car make 86 revolutions as the car reduces its speed uniformly from 90.0 km/h to 58.0 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. rad/s? Submit My Answers Give Up 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 Value Units Submit My Answers Give Up

Explanation / Answer

Here,

theta = 86 rev = 540.3 radians

intial velocity , ui = 90 km/hr

ui = 25 m/s

initial angular speed, wi = 25/(0.84/2) = 61 rad/s

final velocity , uf = 58 km/hr = 16.11 m/s

final angular speed , wf = 16.11/(0.84/2) = 38.4 rad/s

part a) let the angular acceleration is a

Using third equation of motion

wf^2 - wi^2 = 2 * theta * a

38.4^2 - 61^2 = 2 * 540.3 * a

solving for a

a = -2.08 rad/s^2

the angular acceleration is -2.08 rad/s^2

part b)

let the time taken is t

t = wf/a

t = 38.4/2.08

t = 18.5 s

the time taken is 18.5 s

part C)

acceleration , at = 2.08 * (0.84/2) = 0.87 m/s^2

for the total distance
25^2 = 2 * 0.87 * d

d = 357 m

the stopping distance is 357 m

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---