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Rates and motion question (1 Viewer)

aqwerty13402

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How do I calclate both of the things in the title?

Is average velocity y2-y1/x2-x1
and speed is just distance over time?

im confused and i have an exam on tuesday pleease help
 

aqwerty13402

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Can someone please tell me how to do the highlighted part of C? My teacher doesn't know either
1716514821646.png
There is no way to get rid of the constant, so I don't know how the answers are getting a single number?

The stationary point is the maximum distance to the right. But i have no idea how to find the maximum distance to the left. It makes sense to me why they said its the boundary (10 seconds) since its velocity has been negative for the longest. I just dont get how they subbed it back into the displacement function when we dont have the constant C?
 

liamkk112

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Can someone please tell me how to do the highlighted part of C? My teacher doesn't know either
View attachment 43253
There is no way to get rid of the constant, so I don't know how the answers are getting a single number?

The stationary point is the maximum distance to the right. But i have no idea how to find the maximum distance to the left. It makes sense to me why they said its the boundary (10 seconds) since its velocity has been negative for the longest. I just dont get how they subbed it back into the displacement function when we dont have the constant C?
doesn't matter about C, it's only about how far it is away from the initial position. for example:
we can just take the initial position as 0, so x(0) = 0 = C
so now x(t) = 16t-2t^2 = 2t(8-t)
now sketching this, we see that the maximum distance to the right is 32 cm from the intial position (vertex of the parabola at t = 4), and the maximum to the left is 40cm from the initial position (at t = 10).
it wouldn't really matter if we took x(0) = 10, for example. now
x(t) = 10+16t-2t^2 = 2(5+8t-t^2)
again the maximum here is at the vertex of t = 4s, which has a corresponding x(t) of 42. so how far away is it from the initial position of x = 10? 32cm, same as we got before. so u can just pick any value for the initial position, u just need to make sure that the maximum distances u give are relative to that initial position
 

aqwerty13402

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doesn't matter about C, it's only about how far it is away from the initial position. for example:
we can just take the initial position as 0, so x(0) = 0 = C
so now x(t) = 16t-2t^2 = 2t(8-t)
now sketching this, we see that the maximum distance to the right is 32 cm from the intial position (vertex of the parabola at t = 4), and the maximum to the left is 40cm from the initial position (at t = 10).
it wouldn't really matter if we took x(0) = 10, for example. now
x(t) = 10+16t-2t^2 = 2(5+8t-t^2)
again the maximum here is at the vertex of t = 4s, which has a corresponding x(t) of 42. so how far away is it from the initial position of x = 10? 32cm, same as we got before. so u can just pick any value for the initial position, u just need to make sure that the maximum distances u give are relative to that initial position
THANKS!!!!
 

aqwerty13402

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Hey guys sorry, i have an exam tomorrow.

Its a release task so this question could be in the exam:
1716809067719.png

Could someone please do ii and iii for me? I have working out that ive done, thats given me the answers but idk how its right.fuck me dead
 

aqwerty13402

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I AM SO SORRT FOR SPAMMING THREADS, I WILL DELETE THEM AFTER MY EXAM TMRW.

1716812875387.png
For a), i have found out that it is maximised when t = 6. BUt what the fuck do i sub t=6 into to get the flow rate? Idk if the answers are wrong, but they say 1200 m^3 per month???

I CANT DO THIS ANYMORE

also i swear i have done this before, but i didnt reailse you had to find the flow rate, i was only finding t = 6
 

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Aeonium

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I AM SO SORRT FOR SPAMMING THREADS, I WILL DELETE THEM AFTER MY EXAM TMRW.
View attachment 43276

For a), i have found out that it is maximised when t = 6. BUt what the fuck do i sub t=6 into to get the flow rate? Idk if the answers are wrong, but they say 1200 m^3 per month???

I CANT DO THIS ANYMORE

also i swear i have done this before, but i didnt reailse you had to find the flow rate, i was only finding t = 6
do you sub it into dw/dt (1.2- \cos^2 \frac \pi {12} t) thing ? logically the maximum flow is 1200 because the cos^2 term should be minimised
 

aqwerty13402

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do you sub it into dw/dt (1.2- \cos^2 \frac \pi {12} t) thing ? logically the maximum flow is 1200 because the cos^2 term should be minimised
also if u could help with b and c that'd be appreciated heaps. SOrry about this,
 

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