Question

A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. if the radius of a

tire is 29 cm, find the number of revolutions the tire makes during this motion, assuming that no

slipping occurs.​

Answer 1

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Answer 2

Answer:

54.36 revolutions

Step-by-step explanation:

Initial velocity of car, u = 0

Final velocity of car, v = 22m/s

Time = 9 seconds

v=u+at

22=0+a×9

a=(22/9)m/s²

And s=ut+1/2at²

⇒s=at²/2

s=22×9²/9×2               (a=(22/9)m/s²)

Since diameter of the tire is 58cm or 0.58m its circumference would be πr²=3.14×0.58

=1.8212m

So, in one rotation it covers 1.8212m, therefore it will make

99/1.8212 = 54.36 revolutions

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