Finding acceleration from distance and time

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To find the bullet's acceleration as it travels through the rifle's barrel, the key is to use the kinematic equation that relates distance, time, and acceleration: x = 1/2 a t^2. Given that the bullet travels a distance L = 1.02 m in time t = 0.16 s, the equation can be rearranged to solve for acceleration a. The expression for acceleration becomes a = 2L/t^2. This approach utilizes the fact that the initial velocity is zero, simplifying the calculations. Understanding these kinematic principles is essential for solving the problem correctly.
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Homework Statement


A bullet with a mass of m=15.5 g is shot out of a rifle that has a length of L=1.02 m. The bullet spends t=0.16 s in the barrel.

Write an expression, in terms of given quantities, for the magnitude of the bullet's acceleration, a, as it travels through the rifle's barrel. You may assume the acceleration is constant throughout the motion.

So I'm supposed to write an acceleration equation using L and t but everything I try it says is wrong and the only hints are to use the given information and that there's a 2 somewhere in the numerator.

Homework Equations


I'm trying to find the equation. But I assume that V= d/t is used somehow and then a=v(final)+v(initial) / t is also used

The Attempt at a Solution


I tried just putting d/t (or L/t) where v goes in the acceleration equation but it says there's a 2 somewhere in the numerator that is also somehow related to L. So I tried squaring L and also L/t but neither of those was right and I also tried putting (2(L/t))/t which was also wrong. I only have one guess left before I just get a 0 on this part of the problem and the rest of the problem requires this formula (I tried to figure out the rest without doing this part but couldn't).
 
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In an uniform accelerated motion the space traveled by a particle at any time is given by x(t)=1/2at^2

With the initial condition: V(initial)= 0 and x(initial)=0
 
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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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