How Is Average Acceleration Calculated for an Arrow on a Bow?

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To calculate the average acceleration of an arrow on a bow, the displacement is 75 cm and the final velocity is 75 m/s. The average acceleration can be determined using the kinematic equation, which states that average acceleration is the change in velocity divided by the change in time. Since time is not directly provided, it can be calculated by assuming a linear acceleration, leading to the formula a = (final velocity - initial velocity) / time. The discussion emphasizes the importance of using kinematic equations to solve for average acceleration when time is unknown. Understanding these principles is crucial for accurately determining the arrow's acceleration.
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Homework Statement


I will just try to post the whole questions, maybe my idiot eyes missed something...

an arrow is accelerated for a displacement of 75cm [fwd]
while it is on the bow, if the arrow leaves the bow at a velocity of 75m/s [fwd]
what is it's average acceleration while it's on the bow


Homework Equations


t = displacement/ velocity
a = velocity/time


The Attempt at a Solution



75 cm -> .75
0.75 / 75 = 0.01s
disp = 75m

t = 75/0.75 = 100s
velocity = 75m/s

a = velocity / time
time = displacement + velocity / velocity
acceleration = velocity/(disp/velocity) = vecloity / 0.01 = 7500 WTH??
 
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The arrow leaves the bow with a velocity of 75 m/s, but its initial velocity = 0 m/s
 
ya it should be, according to logic I guess lol
just I've been having a lot of trouble trying to figure out the average acceleration
 
Combine said:
ya it should be, according to logic I guess lol
just I've been having a lot of trouble trying to figure out the average acceleration

Average acceleration is the change in velocity divided by the change in time. You are not given the time. So you must assume a linear variation of acceleration while the arrow is in contact with the bowstring (it behaves like a spring), in which case you can use the standard kinematic equations to solve for the average acceleration.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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