Mlopez14
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Could anyone give me a bit of direction for the following question, please?
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RDKGames
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(Original post by Mlopez14)
Could anyone give me a bit of direction for the following question, please?
In the first part of motion, when t<10, what is the velocity in terms of a at t=10 ?

In the second part of motion, what is the velocity when t= 10 ?

Equate them, they must be equal
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Mlopez14
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(Original post by RDKGames)
In the first part of motion, when t<10, what is the velocity in terms of a at t=10 ?

In the second part of motion, what is the velocity when t= 10 ?

Equate them, they must be equal
I already had an idea for that second part, it is the first part that bothers me. I was going to use the equation but isn't it only for 10 < t < 20
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RDKGames
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(Original post by Mlopez14)
I already had an idea for that second part, it is the first part that bothers me. I was going to use the equation but isn't it only for 10 < t < 20
When I say 'part of motion' I am referring to the fact that the particle has two different parts of motion; in the first part, it has constant aceeleration, and in the second one it no longer has constant acceleration.

I was not referring to parts (a) and (b) of the question it self -- because from what you've just said it seems to me this is what you inferred.

Anyway, v = \dfrac{800}{t^2} - 2 is for 10 \leq t \leq 20, so of course you can use t=10 here.
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Mlopez14
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(Original post by RDKGames)
When I say 'part of motion' I am referring to the fact that the particle has two different parts of motion; in the first part, it has constant aceeleration, and in the second one it no longer has constant acceleration.

I was not referring to parts (a) and (b) of the question it self -- because from what you've just said it seems to me this is what you inferred.

Anyway, v = \dfrac{800}{t^2} - 2 is for 10 \leq t \leq 20, so of course you can use t=10 here.
I was using your logic actually but when the particle is in constant acceleration, so when t < 10, can I still use v = \dfrac{800}{t^2} - 2 is for 10 \leq t \leq 20?
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RDKGames
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(Original post by Mlopez14)
I was using your logic actually but when the particle is in constant acceleration, so when t < 10, can I still use v = \dfrac{800}{t^2} - 2 is for 10 \leq t \leq 20?
No you cant use that, and it should be pretty obvious that if you differentiate this you wont get a constant acceleration so using it for motion when 0<t<10 would be in contradiction to what the question tells you.

You need to construct your own velocity equation based on the information and use that for when 0<t<10
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Mlopez14
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(Original post by RDKGames)
No you cant use that, and it should be pretty obvious that if you differentiate this you wont get a constant acceleration so using it for motion when 0<t<10 would be in contradiction to what the question tells you.

You need to construct your own velocity equation based on the information and use that for when 0<t<10
Ah ok, it makes sense, thank you.
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Mlopez14
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Thanks to you, I managed to complete the first two part of the questions. However, I'm stuck for part c. I integrated v = \dfrac{800}{t^2} - 2 to become v = \dfrac{800}{t} - 2t + c . But I am not sure how I have to proceed from now.
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ThiagoBrigido
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(Original post by Mlopez14)
Thanks to you, I managed to complete the first two part of the questions. However, I'm stuck for part c. I integrated v = \dfrac{800}{t^2} - 2 to become v = \dfrac{800}{t} - 2t + c . But I am not sure how I have to proceed from now.
Only use the constant c when there is no limits.
Part c, refers to the displacement of P from 10s to 20s. Remember SuVAt (constant acceleration is the key word). The question has provided you with the velocity, therefore when integrating the velocity you obtain the displacement S=[ - 800/t - 2t ] to be evaluated from 10s to 20s .
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