Help me in this C34 math problem please

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JacobBob
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https://qualifications.pearson.com/c...e_20150126.pdf Q13d
Why isn't the answer to 13d =4*b(ii) ?

And how do I solve it ?
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mqb2766
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For 13d) At t=0, the position is not vertically dowm, so 13bii) is not a half period.
In the previous question parts, you should have something like
Rcos(30t - alpha)
What value of t causes rhe sinusoidal term to return to the initial position of -alphs?

(Original post by JacobBob)
https://qualifications.pearson.com/c...e_20150126.pdf Q12d
Why isn't the answer to 13d =4*b(ii) ?

And how do I solve it ?
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JacobBob
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(Original post by mqb2766)
For 13d) At t=0, the position is not vertically dowm, so 13bii) is not a half period.
In the previous question parts, you should have something like
Rcos(30t - alpha)
What value of t causes rhe sinusoidal term to return to the initial position of -alphs?
Ok I'm not sure I understood what you said correctly, but what I did from what I understood, is find the t at which H is first = 0 then the t at which it ='s 0 for the 2nd time (t2) then I subtracted t2-t1 and multiplied by 2 to get the time for 2 revolutions but I got 12 minutes when the answer in the ms is 24
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mqb2766
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(Original post by JacobBob)
Ok I'm not sure I understood what you said correctly, but what I did from what I understood, is find the t at which H is first = 0 then the t at which it ='s 0 for the 2nd time (t2) then I subtracted t2-t1 and multiplied by 2 to get the time for 2 revolutions but I got 12 minutes when the answer in the ms is 24
Im not too sure exactly what you have dome, but do you understand why its not 4*b(ii)? If the position at t=0 was vertically down, that that would be correct, but it isn't.

You want to find the time to complete one revolution (an additional 360 degrees) and then double it. You have something like
Rcos(30t + alpha)
At t=0, this is Rcos(alpha). What t gives
Rcos(360 + alpha)
You've already basically worked it out. Then double it as you want two revolutions.
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JacobBob
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(Original post by mqb2766)
Im not too sure exactly what you have dome, but do you understand why its not 4*b(ii)? If the position at t=0 was vertically down, that that would be correct, but it isn't.

You want to find the time to complete one revolution (an additional 360 degrees) and then double it. You have something like
Rcos(30t + alpha)
At t=0, this is Rcos(alpha). What t gives
Rcos(360 + alpha)
You've already basically worked it out. Then double it as you want two revolutions.
I get why 4*bii isn't correct but

Are you adding 360 because 1 revolution is 360 or because, with cos, to get to the same y value again (aka the period) you add 360 ?
Last edited by JacobBob; 1 year ago
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mqb2766
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1 rotation is 360.
You could add 720 (two roations) or double the 360 answer. Same thing.
(Original post by JacobBob)
I get why 4*bii isn't correct but

Are you adding 360 because 1 revolution is 360 or because, with cos, to get to the same y value again (aka the period) you add 360 ?
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JacobBob
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(Original post by mqb2766)
1 rotation is 360.
You could add 720 (two roations) or double the 360 answer. Same thing.
You don't understand my question but I really need to understand how to solve this so I'm not gonna give up just yet.
Here's what I understand ( I understood the following based on what you said):
When t=0 , H=2m
We want to know how much time it takes this passenger to get back to 2m
we know that cos 16.7 gives the answer 2m so what is the next angle that gives 2m? 16.7+360 (p.s alpha is 16.7)
360/30t =12
12*2=24

This is what I have come to, but you said 360 is 1 rotation , while I know that to be true, I don't see how/where you used that ? I did not add 360 to to 16.7 because 360 is the angle of 1 rotation, but because, if I were to sketch a cos graph, the values that would give me H=2 would be 360 degrees apart.

What have I missed here ? What don't I understand ? What is incorrect with my thinking ?
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mqb2766
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In post 4 I do add alpha. See below.
(Original post by JacobBob)
You don't understand my question but I really need to understand how to solve this so I'm not gonna give up just yet.
Here's what I understand ( I understood the following based on what you said):
When t=0 , H=2m
We want to know how much time it takes this passenger to get back to 2m
we know that cos 16.7 gives the answer 2m so what is the next angle that gives 2m? 16.7+360 (p.s alpha is 16.7)
360/30t =12
12*2=24

This is what I have come to, but you said 360 is 1 rotation , while I know that to be true, I don't see how/where you used that ? I did not add 360 to to 16.7 because 360 is the angle of 1 rotation, but because, if I were to sketch a cos graph, the values that would give me H=2 would be 360 degrees apart.

What have I missed here ? What don't I understand ? What is incorrect with my thinking ?
(Original post by mqb2766)
You want to find the time to complete one revolution (an additional 360 degrees) and then double it. You have something like
Rcos(30t + alpha)
At t=0, this is Rcos(alpha). What t gives
Rcos(360 + alpha)
You've already basically worked it out. Then double it as you want two revolutions.
So its an additional 360 or 720 degrees, so for 1 rotation
30t = 360
....
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JacobBob
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(Original post by mqb2766)
In post 4 I do add alpha. See below.




So its an additional 360 or 720 degrees, so for 1 rotation
30t = 360
....
Why aren't you answering my question directly (not trying to be rude)

I still don't know if what I understand on why we add 360 and 720 is correct ...
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mqb2766
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TBH I don't fully understand what you're misunderstanding and I'm trying to help you get to the answer / understanding. Try and understand why one completre revolution must satisfy
30t = 360

If you could clearly write down on paper what you think the solution is and upload it, Ill try and comment on that. Please include any working from previous steps if necessary.

(Original post by JacobBob)
Why aren't you answering my question directly (not trying to be rude)

I still don't know if what I understand on why we add 360 and 720 is correct ...
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