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.
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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
HOPE IT HELPS YOU DEAR FRIEND
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