A 350 g frisbee moving at 26 m/s is stopped by a student when he catches it, exerting a constant force of 390 N in the opposite direction. How long does it take him to stop the frisbee? Also, how far does the frisbee travel before stopping?
Well \[F=ma \rightarrow a=\frac{ F }{ m } \]\[a=\frac{ \Delta v }{ t }\rightarrow t=\frac{ \Delta v }{ a }=\frac{ m(v _{f}-v _{i}) }{ F }=\frac{ 0.350kg(0 \frac{ m }{ s } -26\frac{ m }{ s })}{ -390N }\]
The equation above is perfect for solving for the time taken to stop the frisbee. The time taken to stop the frisbee can also be thought of as the amount of time needed to reduce the velocity of the frisbee from its original velocity to 0. If we use Newton's Law of F=ma, we can solve for acceleration and substitute the expression into the kinetics equation for acceleration as a change in velocity over time.
Nevermind, I figured it out. ∆t=(mvf-mvi)/F and ∆x=.5(vi+vf)∆t
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