Finding Constant Acceleration of Train: Liz's Homework Solution

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To find the constant acceleration of the train, Liz calculated the velocity of each car based on their lengths and the time taken to pass. She used the equation change in velocity = a(t) but arrived at an acceleration of 0.8 m/s², which differs from the book's answer of 1.6 m/s². A participant pointed out that using v = s/t only applies to constant velocity, suggesting Liz derive two equations for acceleration and initial velocity to eliminate the initial velocity and solve for acceleration. The discussion highlights the importance of correctly applying physics equations in kinematics. Ultimately, understanding the relationship between velocity, time, and acceleration is crucial for solving such problems accurately.
atypical
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Homework Statement


Liz rushes down onto a subway platform to find her train already
departing. She stops and watches the cars go by. Each
car is 8.60 m long. The first moves past her in 1.50 s and the
second in 1.10 s. Find the constant acceleration of the train.


Homework Equations


change in velocity=a(t)


The Attempt at a Solution


My answer is different from the book. I found the velocity of each cart given the length and time to pass. From there I took the difference of the two velocities to get my change in velocity.
Then, from the equation i found the total time from both cars to pass then divided to solve for a.

change in velocity=a(t)
therefore: 2.11m/s=a(1.5s+1.1s)
a=0.8m/s^2
Book is getting 1.6m/s^2
...so I am off by a product of 2?
 
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hi atypical! :smile:

(try using the X2 icon just above the Reply box :wink:)
atypical said:
I found the velocity of each cart given the length and time to pass.

nooo … v = s/t only works for constant velocity …

call the initial velocity v, and get two equations for a and v, from which you can eliminate v to find a :smile:
 
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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