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19 votes
19 votes
A car accelerates uniformly from rest and

reaches a speed of 32.1 m/s in 13.8 s. The
diameter of a tire is 84.7 cm.
Find the number of revolutions the tire
makes during this motion, assuming no slipping.
Answer in units of rev.

User Walkman
by
3.1k points

1 Answer

16 votes
16 votes

From the information given,

initial speed, u = 0 because the car started from rest

Final speed, v = 32.1 m/s

time, t = 13.8

We would find the distance covered by the car, d by applying one of Newton's formula of motion which is expressed as

d = 1/2(u + v)t

By substituting the values into this formula, we have

d = 1/2(0 + 32.1)13.8

d = 221.49 m

Recall,

d = θR

where

θ is the angular displacement

R is the radius

From the information given,

R = 84.7 cm

We would convert from cm to m.

recall,

100 cm = 1m

84.7 cm = 84.7/100 = 0.847m

Substituting R = 0.847 and d = 221.49 into the formula, it becomes

221.49 = 0.847θ

Dividing both sides of the equation by 0.332, we have

θ = 221.49/0.847

θ = 261.5 radians

1 radian = 0.159 revolutions

261.5 radians = 261.5 x 0.159

= 41.6 revolutions

Number of revolutions = 41.6

User OdinX
by
2.9k points