Question

A car 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.

Answer 1

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

$v=u+(a \times t)$

$33.5=0+(a \times 11.9)$

$11.9 \times a=33.5$

$a=\frac{33.5}{11.9}=2.815 \mathrm{m} / \mathrm{s}^{2}$

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

$s=(u \times t)+\left(\frac{1}{2} \times a \times t^{2}\right)$

$s=0+\left(0.5 \times 2.815 \times 11.9^{2}\right)$

$s=0.5 \times 2.815 \times 141.61=199.31 m$

Now as in each revolution a wheel rolling without slipping covers a distance equal to its periphery ($2 \pi R$), if the total numbers of revolution of each tire be n, then we have

$s=n \times 2 \pi R$

Given diameter, d = 37.5 cm = 0.375 m

Radius of the wheel, R =  $\frac{d}{2}=\frac{0.375}{2}=0.1875 \mathrm{m}$

Total numbers of revolution,

$n=\frac{s}{2 \pi R}=\frac{199.31}{2 \times 3.14 \times 0.1875}=\frac{199.31}{1.1775}=169.265 \text { revolution }$