Question
A motorcyclist is traveling along a road and accelerates for 4.45 s to pass another cyclist. The angular acceleration of each wheel is +6.35 rad/s2, and, just after passing, the angular velocity of each wheel is +75.6 rad/s, where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?
Answers
v= 75.6
v=u+at
at= 6.35 x 4.45= 28.26
75.6 - 28.26 = u = 47.34
v^2=u^2+2as
75.6^2 = 47.34^2+(2 x 6.35 x s)
5715.36-2241.08=12.7 x s
3474.27/12.7= s (answer is in rad)
v=u+at
at= 6.35 x 4.45= 28.26
75.6 - 28.26 = u = 47.34
v^2=u^2+2as
75.6^2 = 47.34^2+(2 x 6.35 x s)
5715.36-2241.08=12.7 x s
3474.27/12.7= s (answer is in rad)
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