Calculating Time for Arrow to Stop Moving Respect to Block

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To determine how long it takes for the arrow to stop moving relative to the block, the equation v = Vinitial + acceleration * time is used, with the arrow's deceleration at -1550 m/s² and the block's acceleration at 450 m/s². The key is to set the velocities of the arrow and block equal at the moment the arrow comes to rest relative to the block. By substituting the known values into the equation, the correct time can be calculated. The discussion highlights the importance of understanding relative motion and the application of kinematic equations. The problem was resolved successfully after correcting algebraic errors.
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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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