Asked by Too Hard!!!!
A stuntman drives a dirt bike on a curved track with a radius of 17.0 m. If he starts from rest and accelerates at 1.74 m/s^2, at what time will the magnitude of the total acceleration of the bike be 7.04 m/s^2?
Answers
Answered by
Damon
inward Ac = v^2/r
tangential a = dv/dt
7.04 =sqrt[a^2 + (v^2/r)^2 ]
49.56 = 3.03 + v^4/289
v = a t = 1.74 t
v^4 = 9.17 t^4
so
46.53 = 9.17 t^4/289
t^4 = 1466
t = 38.3 seconds
Check my arithmetic!!!
tangential a = dv/dt
7.04 =sqrt[a^2 + (v^2/r)^2 ]
49.56 = 3.03 + v^4/289
v = a t = 1.74 t
v^4 = 9.17 t^4
so
46.53 = 9.17 t^4/289
t^4 = 1466
t = 38.3 seconds
Check my arithmetic!!!
Answered by
Nickelous
The WindSeeker ride at Canada's Wonderland lifts riders up to 300 feet into the air while spinning them around in a 12.2 m radius circle. The WindSeeker starts its circular motion from rest with a constant angular acceleration of 0.0174 rad/s2. It maintains that angular acceleration for 48.1 s, after which the angular speed remains constant for the rest of the ride - another 120. s. What is the top speed of the ride?
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