One dimensional motion problems

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Runner A starts 4 miles west of a flagpole and runs east at 6 mph, while Runner B starts 3 miles east and runs west at 5 mph. To find their meeting point, their displacements are set equal, resulting in a calculation that shows they meet 0.2 miles west of the flagpole. For the train problem, with an initial speed of 20 m/s and a deceleration of -1.0 m/s², the time to stop is calculated, and the distance traveled during the braking period is determined using the appropriate kinematic equations. Both problems illustrate the application of one-dimensional motion equations to solve for distances and meeting points.
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Homework Statement


1) Runner A is initially 4.0 mi west of a flagpole and is running with a constant velocity of 6.0 mi/h due east. Runner B is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when they meet?



2) A train is traveling down a straight track at 20 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s^2 as long as the train is in motion. How far does the train move during a 40 s time interval starting at the instant the brakes are applied?


Homework Equations


1) V=Vi + at
2) x=Vit + 1/2at^2
3) V^2=Vi^2 + 2ax
4) a=V/t
5) v=x/t



The Attempt at a Solution

 
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1) Can you find the displacement (with respect to the flagpole) of each runner at any given time? When they meet, what must their displacement be?

2) How long does it take for the train to stop with the given acceleration? Thus how far would it have travelled?
 
Set up an equation with respect to the flagpole being zero. If runner a is 4mi west (-4miles) and running 6mi/h east (+6mi/h) and runner b is 3 miles east (+3miles) running at 5mi/h west (-5mi/hr), then you can set them equal to each other to find when they meet. -4+6x=3-5x then x=.64 Then plug .64 back into one side of the equation (3-5(.64))= -.2 miles (.2 miles west). Thats how I got the right answer for my webassign.
 
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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