A motorcycle stunt rider rides off the edge of a cliff. Just at the edge, his velocity is horizontal with magnitude 9.0m/s. Find the motorcycle’s distance from the edge of the cliff and velocity after 0.5s.
A
Distance =3494m, Velocity =√106m/s
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B
Distance =√3492m, Velocity =√106m/s
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C
Distance =√3494m, Velocity =106m/s
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D
Distance =√3494m, Velocity =√106m/s
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Solution
The correct option is D Distance =√3494m, Velocity =√106m/s
At t=0.50s, the x and y-coordinates are x=v0t=(9.0m/s)(0.50s)=4.5m y=−12gt2=−12(10m/s2)(0.50s)2=−54m
The negative value of y shows that the motorcycle is below its starting point.
The motorcycle’s distance from the origin at this time r=√x2+y2=√(92)2+(54)2=√3494m.
The components of velocity at this time are vx=v0=9m/s,vy=−gt=(−10m/s)(0.50s)=−5m/s
The speed (magnitude of velocity) at this time is v=√v2x+v2y=√(9m/s2)+(−5m/s2)=√106m/s