How to solve SUVAT problem without using relative velocity

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To solve the problem of calculating the time taken for particle A to catch up with particle B, one method involves writing separate position equations for each particle and equating them to find the meeting point. This simultaneous equation approach can be advantageous, especially when dealing with more complex motions or separate accelerations. However, using relative velocity is also valid and can simplify the process in many cases. Both methods can be interconverted through algebraic manipulation, and the choice may depend on the specific problem's complexity. Ultimately, both approaches yield the same result, allowing flexibility in problem-solving strategies.
rollcast
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Homework Statement


Calculate the time taken for particle A to catch up with particle B.
See attachment for variables.

I worked out the answer by finding the relative velocity of A wrt B but apparently there's another method I should have/could have used to solve it?


Homework Equations



SUVAT equations of motion


The Attempt at a Solution



See attachment.

Thanks
AL
 

Attachments

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You could write separate expressions for the positions of each particle, then equate them to find when the positions are the same.
 
So call the final displacement when they meet, x, get 2 equations and then solve them as simultaneous equations?

Is there any advantage to that approach compared to just working out the relative velocity?
 
rollcast said:
So call the final displacement when they meet, x, get 2 equations and then solve them as simultaneous equations?

Is there any advantage to that approach compared to just working out the relative velocity?

Either approach could be transformed into the other via suitable algebraic slight of hand. The separate equation approach has the advantage that it may be more straightforward when the particle motions are more complicated. For example, suppose that both particles also had separate accelerations?
 
gneill said:
Either approach could be transformed into the other via suitable algebraic slight of hand. The separate equation approach has the advantage that it may be more straightforward when the particle motions are more complicated. For example, suppose that both particles also had separate accelerations?

But the relative approach will still work with accelerations?

Do you just work out the acceleration of one relative to the other or do you have to use v=u+at for both and then then subtract the 2?
 
rollcast said:
But the relative approach will still work with accelerations?

Do you just work out the acceleration of one relative to the other or do you have to use v=u+at for both and then then subtract the 2?

It amounts to the same thing.
 
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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