a car accelerate uniformly from rest and reach a speed of 22.0 m/s in 9.00 s . if the diameter of the tire is 58.0 cm , find (a) the number of revaloution the tire makes during this motion assuming that no slipping occurs. (b) what is the final rotational speed of a tire in revalotation per second ?
Answers
Answer:
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Answer:
Given: Initial velocity = 0 (starts from rest)
Final velocity = 22 m/s
time, t = 9 s
Diameter of a tire = 58 cm
Find: (a) The number of revolutions the tire makes during the motion
(b) The final angular apeed of the tire.
Step 2
(a) To calculate number of revolutions, first calculate acceleration, as:
Vf=Vi+at
a=(Vf-Vi)÷t
a=(22-0)÷9
a=2•44 m per second square
NOW,
a=ar
r=58÷2 cm=0•58÷2=0•29m
a=2.44/0.29
a=8.41379