Problem 2: During a very quick stop, a car decelerates at 7.2 m/s 2 . Assume the

Question

Problem 2: During a very quick stop, a car decelerates at 7.2 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).

Part (a) What is the angular acceleration of its 0.275 m-radius tires, assuming they do not slip on the pavement in rad/s2?

Part (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 98 rad/s ?

Part (c) How long does the car take to stop completely in seconds?

Part (d) What distance does the car travel in this time in meters?

Part (e) What was the car

Problem 2: During a very quick stop, a car decelerates at 7.2 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).

Part (a) What is the angular acceleration of its 0.275 m-radius tires, assuming they do not slip on the pavement in rad/s2?

Part (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 98 rad/s ?

Part (c) How long does the car take to stop completely in seconds?

Part (d) What distance does the car travel in this time in meters?

Part (e) What was the car

Explanation / Answer

a) Angular accelaration = a/R = -7.2 /0.275 =-26.18 rad/s2

b)

v = u - at

or, 0 = 98 - 26.18t

or, t = 98/26.18 = 3.74 s

Now, theta = ut - 1/2at2

or, theta = 98 x 3.74 - 1/2 x 26.18 x 3.742

or, theta = 183.42 rads

2pi rads make one revolution.

Thus, 183.42 rads make 183.42/ 2pi revolutions = 29.2 revolutions.

c)

v = u - at

or, 0 = 98 - 26.18t

or, t = 98/26.18 = 3.74 s

d) In one revolution, distance covered = 2piR

Thus, in 29.2 revolutions, distance covered by car = 2piR x 29.2 = 2 x pi x 0.275 x 29.2 = 50.45 m

e) Distance covered by car = 50.45 m in 3.74s with decelaration of 7.2m/s2

Thus, v = u - at

0 = u - 7.2 x 3.74

or, u = 7.2 x 3.74 = 26.928 m/s

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