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Bows and Arrows

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An arrow while being shot from a bow was accelerated over a distance of 2.0 ft. If its speed at the moment it left the bow was 200 ft/sec what was the average acceleration imparted by the bow? Justify any assumptions you need to make.

Ok so I know [itex] v_{0} = 200 \frac{ft}{sec} [/itex]. Also [itex] t = \frac{1}{100} [/itex] second. So would I use the equation [itex] x = x_{0}+v_{x}_{0}t + \frac{1}{2}a_{x}t^{2} [/itex]? Or [itex] 2 = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2} [/itex] or [itex] 2 = 2 + \frac{1}{2}a_{x}(\frac{1}{100})^{2} [/itex]? I dont think this makes any sense. Maybe I need to make some assumptions?

Thanks
 
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Answers and Replies

Average acceleration is just change in velocity over change in time. You already have everything you need, but I'll give you a hint with your assumption: What's your starting speed?
 
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Indeed, [itex] \overline a = \frac{v_{2}-v_{1}}{t_{2}-t_{1}} = \frac{\Delta v}{\Delta t} [/itex]. So I assume that the starting speed is 0. So we have [itex] \frac{200}{\frac{1}{100}} = 20,000 \frac{ft}{sec^{2}} [/itex]. Is this correct?
 
Yup, no need for messy position formulas. If you really want to test your knowledge of concepts(and if you've covered this material yet), what's the average force exerted by bow onto the arrow? What's the work done by the arrow? Where does the arrow land, and how long is it in flight?
 

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