Kinematics - Sled Homework Solution

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The sled homework problem involves calculating average velocity, acceleration, speed at a specific point, time taken to reach that point, and distance traveled in subsequent intervals. The average velocities for each 2-second interval were calculated as 5.6 m/s, 7.2 m/s, and 8.8 m/s. The constant acceleration was determined to be 0.80 m/s² using two velocity points. There was confusion regarding the sled's speed at the 14.4 m mark, with two different methods yielding slightly different results of 4.8 m/s and 4.0 m/s. The discussion emphasizes the importance of using the correct kinematic equations to resolve these discrepancies.
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Homework Statement


A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.

a) What is the magnitude of the average velocity of the sled during each of the 2.00-s intervals after passing the 14.4-m point?
b) What is the acceleration of the sled?
c) What is the speed of the sled when it passes the 14.4-m point?
d) How much time did it take to go from the top to the 14.4-m point?
e) How far did the sled go during the first second after passing the 14.4-m point?



Homework Equations


x=x_0+v_0*t+1/2at^2
Other kinematic equations


The Attempt at a Solution


So the first thing I did was drew a picture with labels to get an idea of what the problem looked like to get a better idea of it.
[Part A] asks for the average velocity of each 2 second interval so I found those by taking their distance between each and dividing it by the time, so I got 5.6m/s, 7.2m/s and 8.8m/s.
[Part B] asks for acceleration and because it is constant I can take any two points and find the acceleration that way so I took 7.2m/s and 5.6m/s and divided that by 2s to get an acceleration of .80m/s^2.
[Part C] asks for the velocity at the 14.4m point. This one I'm having issues with. I'm sure it's a small mistake I'm overlooking but plugging it into the standard equation 25.6m=14.4m+v_0*2s+.5*.80m/s^2*2^2 yields 4.8m/s. Putting it into the v_0=v_1-at equation however, using v_1=5.6m/s, a=.80m/s^2 and t=2s gives me 4.0m/s. I'm confused as to why these are not the same. If I can get this issue worked out I'll be able to work the other questions easily.

If I've done anything wrong in A or B please point it out to me, thanks.
 
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Basically, v_1 is not the average velocity over the 2 seconds.
 
Alright I think I got it now. Since I don't know the time it took to get to 14.4m I can't find an exact v_1 using an averaged velocity as it isn't't constant so for that I would use the original equation finding v_1 with the constant acceleration, time and distance which would give me an exact number. Thanks.
 
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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