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 can be calculated using the formula for average acceleration, which is the change in velocity divided by the change in time. Assuming the initial speed is 0 ft/sec and the final speed is 200 ft/sec, with a time of 0.01 seconds, the average acceleration is determined to be 20,000 ft/sec². The discussion highlights the importance of making assumptions about initial conditions to simplify calculations. Additionally, it raises further questions about the average force exerted by the bow and the work done by the arrow. Understanding these concepts is essential for analyzing the dynamics of projectile motion.
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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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