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Question

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge, his velocity is horizontal with magnitude 9.0 m/s. Find the motorcycle’s distance from the edge of the cliff and velocity after 0.5 s.

A
Distance =3494 m, Velocity =106 m/s
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B
Distance =3492 m, Velocity =106 m/s
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C
Distance =3494 m, Velocity =106 m/s
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D
Distance =3494 m, Velocity =106 m/s
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Solution

The correct option is D Distance =3494 m, Velocity =106 m/s


At t=0.50 s, the x and y-coordinates are
x=v0t=(9.0 m/s)(0.50 s)=4.5 m
y=12gt2=12(10 m/s2)(0.50 s)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=3494 m.

The components of velocity at this time are
vx=v0=9 m/s,vy=gt=(10 m/s)(0.50 s)=5 m/s
The speed (magnitude of velocity) at this time is
v=v2x+v2y=(9 m/s2)+(5 m/s2)=106 m/s

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