A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 8.91 s.

Question

A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 8.91 s. Assume the diameter of a tire is 57.7 cm.

(a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs.

(b) What is the final angular speed of a tire in revolutions per second?

4 0/1 points | Previous Answers SerPSET9 10.P.020 A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 8.91 s. Assume the diameter of a tire is 57.7 cm (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. Enter a number e total distance traveled by the car in this time interval, how could you determine the total number of revolutions of the tires? rev (b) What is the final angular speed of a tire in revolutions per second? rev/s Need Help?Read It Watch It

Explanation / Answer

Using kinematics equation , we have

v = u + at

21.5 = 0 + a*8.91

a = 2.413 m/s2

Distance covered is given as

d = ut + 1/2at2

d = 0 + 1/2*2.413*8.912

d = 95.782 m

So, Number of revolutions N = 95.782 / 2*pi*0.2885

N = 52.84 revolutions

a = r*alpha

alpha = a/r = 2.413 / 0.2885 = 8.364 rad/s2

w = alpha*t

w = 8.364*8.91

w = 74.522

w = 74.522 / 2*pi

w = 11.86 rev/s

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