A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 11.4 s. The diameter of a tire is 66.5 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.

57.39 rev

Explanation:

From circular motion,

s = rθ ... Equation 1

Where s = distance, r = radius, θ = angular distance.

make θ the subject of the equation

θ = s/r ... Equation 2

Where can look for s using any of the equation of motion

s = (v+u) t/2 ... Equation 3

Where v and u = Final and initial velocity respectively, t = time.

Given: v = 21.5 m/s, u = 0 m/s (at rest), t = 11.4 s

Substitute into equation 3

s = (21.5+0) 11.4/2

s = 122.55 m.

given: r = 66.5/2 = 33.25 cm = 0.3325 m

Substitute into equation 2

θ = 122.55/0.3325

θ = 368.57 rad

θ = (360.57*0.159155) rev

θ = 57.39 rev

58.6886 revolutions

Explanation:

First we need to know the total distance travelled by the car, and we can do that using Torricelli formula:

V2 = Vo2 + 2aDS

V = 21.5

Vo = 0

a = 21.5/11.4 = 1.886

(21.5) ^2 = 2*1.886*DS

DS = 462.25/3.772 = 122.5477 m

For each revolution of the tire, the car moves the circunference of the tire, which is pi*d = 3.14*66.5 = 208.81 cm = 2.0881 m

So, to know the number of revolutions, we divide the total travel distance by the circunference of the tire:

122.5477/2.0881 = 58.6886