Question

A car accelerates uniformly from rest and reaches a speed of 22.5 m/s in 8.96 s. Assume the diameter of a tire is 58.9 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second

Answer 1

Answer:

54.4747649021

38.2003395586 rad/s

Explanation:

Acceleration

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{22.5-0}{8.96}\\\Rightarrow a=2.51116071429\ m/s^2$

Distance covered

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 2.51116071429\times 8.96^2\\\Rightarrow s=100.8\ m$

Number of revolutions

$n=\dfrac{s}{\pi d}\\\Rightarrow n=\dfrac{100.8}{\pi 0.589}\\\Rightarrow n=54.4747649021$

Number of revolutions is 54.4747649021

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{22.5}{0.589}\\\Rightarrow \omega=38.2003395586\ rad/s$

The final angular speed is 38.2003395586 rad/s