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

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Answer:

54.4747649021

38.2003395586 rad/s

Explanation:

Acceleration

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

Distance covered

[tex]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[/tex]

Number of revolutions

[tex]n=\dfrac{s}{\pi d}\\\Rightarrow n=\dfrac{100.8}{\pi 0.589}\\\Rightarrow n=54.4747649021[/tex]

Number of revolutions is 54.4747649021

Angular speed is given by

[tex]\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{22.5}{0.589}\\\Rightarrow \omega=38.2003395586\ rad/s[/tex]

The final angular speed is 38.2003395586 rad/s