Calculating Time for Arrow to Stop Moving Respect to Block

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SUMMARY

The discussion focuses on calculating the time it takes for an arrow, initially traveling at 20.0 m/s, to stop moving relative to a block of styrofoam. The arrow decelerates at 1550 m/s² while the block accelerates at 450 m/s². By applying the equation v = Vinitial + acceleration * time, the participants determined that the velocities of both the arrow and the block must equalize at a certain time, allowing for the calculation of the time variable. The solution was successfully derived through algebraic manipulation of the equations.

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an arrow is fired with a speed of 20.0m/s at at block of styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. during this process the arrow's deceleration has a magnitude of 1550m/s2 and the block's acceleration has a magnitude of 450m/s2. How long does it take for the arrow to stop moving with respect to the block?



Homework Equations


v=vintial + acceleration(t) solve for time



The Attempt at a Solution



i've used this equation but can't figure out how to get time without the positioning of the block from where the arrow was fired.
 
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The arrow has constant velocity during its flight so you do not need the position of the block. After the arrow touches the block, its velocity decreases according the formula for v=Vinitial +acceleration * time, (the acceleration is a negative number here) and the block accelerates according to the same formula.
At some time instant the arrow is in rest with respect to the block. That means that the velocity of the block equals to the velocity of the arrow:

Vi(arrow) + a(arrow)t = Vi(block)+a(block)t.

Plug in the data and solve for t.ehild
 
thanks! it worked out fine. I guess i was doing some wrong algebra before. Thanks for fast response.
 

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