What Is the Average Acceleration of an Arrow Shot from a Bow?

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The average acceleration of an arrow shot from a bow, given an initial speed of 200 ft/sec and a distance of 2.0 ft, is calculated using the formula for average acceleration, \(\overline{a} = \frac{\Delta v}{\Delta t}\). Assuming the starting speed is 0 ft/sec and the time taken is \(\frac{1}{100}\) seconds, the average acceleration is determined to be 20,000 ft/sec². This calculation simplifies the process by avoiding complex position formulas, focusing instead on the change in velocity over time.

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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 v_{0} = 200 \frac{ft}{sec}. Also t = \frac{1}{100} second. So would I use the equation x = x_{0}+v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}? Or 2 = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2} or 2 = 2 + \frac{1}{2}a_{x}(\frac{1}{100})^{2}? I don't think this makes any sense. Maybe I need to make some assumptions?

Thanks
 
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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?
 
Indeed, \overline a = \frac{v_{2}-v_{1}}{t_{2}-t_{1}} = \frac{\Delta v}{\Delta t}. So I assume that the starting speed is 0. So we have \frac{200}{\frac{1}{100}} = 20,000 \frac{ft}{sec^{2}}. 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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