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

54.4747649021

38.2003395586 rad/s

Step-by-step explanation:

Acceleration

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(22.5-0)/(8.96)\\\Rightarrow a=2.51116071429\ m/s^2`

Distance covered

`s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)* 2.51116071429* 8.96^2\\\Rightarrow s=100.8\ m`

Number of revolutions

`n=(s)/(\pi d)\\\Rightarrow n=(100.8)/(\pi 0.589)\\\Rightarrow n=54.4747649021`

Number of revolutions is 54.4747649021

Angular speed is given by

`\omega=(v)/(r)\\\Rightarrow \omega=(22.5)/(0.589)\\\Rightarrow \omega=38.2003395586\ rad/s`

The final angular speed is 38.2003395586 rad/s