Acar accelerates uniformly from rest and reaches a speed of 33.5 m/s in 11.9 s. the diameter of a tire is 37.5 cm.
find the number of revolutions the tire makes during this motion, assuming no slip- ping.

The number of revolutions the tire makes during this motion is 169.265 revolution.

Explanation:

Given:

Car’s initial velocity, u = 0

Car’s final velocity, v = 33.5 m/s

Time required t = 11.9 s

Hence, car’s acceleration ‘a’ can be given by the first equation of motion as

The distance covered by the car during this time is given by the second equation of motion

Now as in each revolution a wheel rolling without slipping covers a distance equal to its periphery (), if the total numbers of revolution of each tire be n, then we have

Given diameter, d = 37.5 cm = 0.375 m

Radius of the wheel, R =  

Total numbers of revolution,

+15-35=-20

going in opposite directions