# Rates and motion question (1 Viewer)

#### aqwerty13402

##### Well-Known Member
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

#### Aeonium

##### Member
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
velocity = (final position - initial position)/time
speed = total distance/time

#### aqwerty13402

##### Well-Known Member
velocity = (final position - initial position)/time
speed = total distance/time
legend

#### aqwerty13402

##### Well-Known Member
Can someone please tell me how to do the highlighted part of C? My teacher doesn't know either

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

##### Well-Known Member
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

##### Well-Known Member
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

##### Well-Known Member
Hey guys sorry, i have an exam tomorrow.

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

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

##### Well-Known Member
WAIT NEVERMIND I SOLVED IT WOOP WOOP

#### harperissleepy

##### Well-Known Member
WAIT NEVERMIND I SOLVED IT WOOP WOOP
YAY GOOD JOB!!! GOOD LUCK ON YOUR EXAM!!

#### aqwerty13402

##### Well-Known Member
I AM SO SORRT FOR SPAMMING THREADS, I WILL DELETE THEM AFTER MY EXAM TMRW.

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

##### Member
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

##### Well-Known Member
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
subbing it into that gives u 9/20

#### aqwerty13402

##### Well-Known Member
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

##### Well-Known Member
GUYS THE EXAM WAS REALLY FUCKING EASY. I NEVER say that about exams, but out of all the qestions they released to us, they picked the easiest ones. Hopefully 90%+