The tires of a car make 62 revolutions as the car reduces its speed uniformly fr

ID: 1353019 • Letter: T

Question

The tires of a car make 62 revolutions as the car reduces its speed uniformly from 88.0 km/h to 64.0 km/h . The tires have a diameter of 0.82 m

Part A

What was the angular acceleration of the tires?

Express your answer using two significant figures.

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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.

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Part C

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

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

= rad/s2

A) v1 = 88 km/h = 88*5/18 = 24.44 m/s

v2 = 64 km/h = 64*5/18 = 17.78 m/s

radius, r = d/2 = 0.82/2 = 0.41m

distance travelled during this time, d = 62*2*pi*0.41

= 159.6 m

tangential acceleration, a_tan = (v2^2 - v1^2)/(2*d)

= (17.78^2 - 24.44^2)/(2*159.6)

= -0.88 m/s^2

angular acceleration, alfa = a_tan/r

= -0.88/0.41

= -2.15 rad/s^2

B) v' = v2 + a*t

t = (v' - v2)/a

= (0 - 17.78)/(-0.88)

= 20.3 s

C) d = (v'^2 - v2^2)/(2*a)

= (0^2 - 17.78^2)/(2*(-0.88))

= 179.6 m

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