How far does arrow penetrate into board

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The discussion revolves around calculating how far an arrow penetrates into a block of Styrofoam after being fired at a speed of 21.0 m/s. The arrow decelerates at 1560 m/s² while the block accelerates at 450 m/s². The user initially struggles with part c of the problem, questioning whether the deceleration is relative to the block or a common reference point. They ultimately find success using the equation x = x0 + 1/2 (v0 + v) t to determine the penetration distance. The conversation highlights the importance of correctly interpreting acceleration references in physics problems.
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[SOLVED] How far does arrow penetrate into board

Homework Statement


An arrow is fired with a speed of 21.0 m/s at a 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 1560 m/s^2 and the block's acceleration has a magnitude of 450 m/s^2.
a)How long does it take for the arrow to stop moving with respect to the block?
b)What is the common speed of the arrow and block when this happens?
c)How far into the block does the arrow penetrate?


Homework Equations


x=Xi+Vi*t+ 1/2at^2
x=0+21m/s (1.04x10^-2 s)+ 1/2 (-1560m/s^2)(1.04*10^-2 s)^2
x=0.2184-0.0843648
x=0.134m



The Attempt at a Solution



I got parts a and b. But part c is giving me problem. I tried using the equation above to get the final position of the arrow but the answer was wrong. Am I using the wrong equation. Or is it a math error?
 
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Is the deceleration of the arrow (-1560 m/s^2) with respect to the block or some common reference point?
 
Unfortunately, the problem doesn't specify that. I'm guessing since the problem gives the deceleration of the arrow while saying "during this process" it might be with respect to the block.
 
if it is wid respect to the block...then use conservation of momentum ...
go ahead..
 
ok. But just out of curiosity, what if it was just with respect to a common point. What would I do then?
 
looking over conservation of momentum, i don't think that's how i need to solve the problem seeing as how I haven't been taught that. Besides that, I have no mass included in the problem.

I'm pretty sure I have to use one of the equations that relate velocity, time, position and acceleration. I used one of them( see above), but got the wrong answer.
 
try using this formula
x = x0 + 1/2 (v0 + v) t
 
that one worked. thanks amarett0!
 
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