Bus and Sports Car Kinematics Problem

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A bus traveling at 15.0 m/s east begins to decelerate at 0.5 m/s² west, while a sports car starts at 6 m/s east with an acceleration of 2.5 m/s² east. The problem involves setting up equations for the distance traveled by both vehicles over time. The key is to equate the two distance equations, as they will be equal when the sports car catches up to the bus. The solution indicates that the sports car will catch the bus in 6 seconds.
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Homework Statement


Question reads: A bus traveling at 15.0m/s [E], started to slow down with an acceleration of 0.5m/s^2[W] as soon as it was abreast of a sports car. At the same instant, the sports car had a velocity of 6m/s [E] and was accelerating at 2.5m/s^2 [E]. At what time does the sports car catch up to the bus?

Homework Equations


vf^2 = vi^2 + 2ad
d = Vit + 1/2at^2


The Attempt at a Solution



d = 15t + 1/2(-0.5)t^2 <-- bus
d = 6t + 1/2(2.5)t^2 <-- car

these are the two equations that i have and will i use these two equations to solve for time ? or distance?

im very confused at this point. ( putting two different object's speed together)

please help me further guide me with this problem.
 
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Hi ya,

The equation you have used is correct - remember, that if the car has caught up with the bus, then d will be the same for both. Hope this helps :)
 


The question sounds so confusing to me :S If the bus and the car are abreast, then the car has already caught up with the bus..

Anyway, usually for this kind of thing, you have 2 equations, use the variable that is in common with both of them, equate them to each other and solve
 


d = 15t + 1/2(-0.5)t^2 <<--- for the bus

d = 6t + 1/2(2.5)t^2 <<-- for the sport car.


Solve the simultaneous equation and you had it.

I hope I'm right @_@.

My answer is 6seconds.
 
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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