Not sure why I can't use v=v0 + at on this problem.

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The discussion revolves around the application of the kinematic equation v = v0 + at to calculate the final velocity of a freight train. The correct final velocity, after accelerating at 0.0400 m/s² for 8 minutes from an initial velocity of 4.00 m/s, is 23.2 m/s. A participant mistakenly calculated a final velocity of 76.8 m/s due to an error in their setup of the equation. The correct calculation involves multiplying the acceleration by the total time in seconds, leading to the accurate result. The thread highlights the importance of careful calculation and understanding of the equations in physics problems.
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Homework Statement



Freight trains can produce only relatively small acceleration and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0400 m/s^2 for 8.00 min, starting with an initial velocity of 4.00 m/s?

Homework Equations



v=v0 + at
x=x0 +v0t +1/2at^2
v^2=v0^2 +2a(x-x0)

The Attempt at a Solution



Why can't i use the first equation I listed to solve this problem? Using that equation, you end up with v = 76.8 m/s

If you first use the second equation, and then the third equation, you end up with v = 23.2 m/s

Thanks in advance!
 
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The first eqn gives the same as the 2nd and third.

v = 76.8 m/s is wrong, 23.2 m/s is right you may have pressed a wrong button on your calculator.

v=v0 + at
v = 4 m/s + (0.04 m/s^2)*(8 min * 60s/min) = 23.2 m/s
 
The first equation is exactly what you're supposed to use. Using the second and third seems roundabout.

But how are you getting 76.8 m/s with the first equation? Can you show your work?
 
Oh goodness, it's been a long weekend haha. I accidentally set it up as 4 m/s TIMES (0.04m/s^2)(480s).

Thanks guys
 
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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