# variable acceleration

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Thread starter 4 weeks ago
#1
I am attaching a question and the book answer. I'm fine up to part d but then I'm not. It says sinkt is always 1 but I don't think it is. When t = 3pi it's -1. Am i missing something subtle?

And then they differentiate a to find max a. I did to and I get t/2 = 0, pi, 2pi and I subbed these into a = 4cos(t/2) and got 4 and -4 so I said it's 4. But why have they taken the route that they have.

I'm missing something but what?
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4 weeks ago
#2
(Original post by maggiehodgson)
I am attaching a question and the book answer. I'm fine up to part d but then I'm not. It says sinkt is always 1 but I don't think it is. When t = 3pi it's -1. Am i missing something subtle?

And then they differentiate a to find max a. I did to and I get t/2 = 0, pi, 2pi and I subbed these into a = 4cos(t/2) and got 4 and -4 so I said it's 4. But why have they taken the route that they have.

I'm missing something but what?
Any chance you could upload larger images? Theyre very blurred.

However it seems to ask for maximum velocity and acceleration, so when sin() and cos()=1. Minumum would be sin() and cos() =-1.
Last edited by mqb2766; 4 weeks ago
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Thread starter 4 weeks ago
#3
(Original post by mqb2766)
Any chance you could upload larger images? Theyre very blurred.

However it seems to ask for maximum acceleration, so when sin()=1. Minumum would be sin()=-1.
In summary: v=2+8sinkt, k is a constant, initial acceleration is 4m/s/s/
a) find the value of k (0.5) done that
b) find in term of Pi the values of t in the interval 0<=t<=4Pi (Pi, 3pi) done that
c) show that 4a^2 = 64 - (v-2)^2
d) find the maximum velocity and the maximum acceleration. At this point, although I got the same answers as in the book I did not understand some of their workings

Here's the book answer
At maximum velocity, a = 0 ms^−2 (agreed)
From part b, this occurs when t = π s and t = 3π s. (agreed)
In both cases, sin1kt=1, so v = 2 + 8 = 10. (disagree sin (3pi/2 does not = 1)
The maximum value of a occurs when da/dat=0
(at this point I differentiated and found t/2 = 0 t 0, pi and 2pi
I then subbed it in to a = 4cos(t/2) and found max acc = 10)

But this is what the book did and I don't know why they dis it that way.
Since a is a multiple of coskt, da/dt is a multiple of sinkt,
so da/dt=0 when sinkt=0 and hence v = 2 + 0 = 2 (from a)
By the result from c, at v=2 we have (why do this and not sub into a = 4cos(t/2)?)
4a^2 = 64(-(2-2)^2
a = 4
The maximum values of velocity and acceleration are 10 ms−1 and 4 ms−2 respectively.

Thanks. Sorry for the rubbish photos. I don't know how to get them to be bigger or clearer.
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4 weeks ago
#4
(Original post by maggiehodgson)
In summary: v=2+8sinkt, k is a constant, initial acceleration is 4m/s/s/
a) find the value of k (0.5) done that
b) find in term of Pi the values of t in the interval 0<=t<=4Pi (Pi, 3pi) done that
c) show that 4a^2 = 64 - (v-2)^2
d) find the maximum velocity and the maximum acceleration. At this point, although I got the same answers as in the book I did not understand some of their workings

Here's the book answer
At maximum velocity, a = 0 ms^−2 (agreed)
From part b, this occurs when t = π s and t = 3π s. (agreed)
In both cases, sin1kt=1, so v = 2 + 8 = 10. (disagree sin (3pi/2 does not = 1)
The maximum value of a occurs when da/dat=0
(at this point I differentiated and found t/2 = 0 t 0, pi and 2pi
I then subbed it in to a = 4cos(t/2) and found max acc = 10)

But this is what the book did and I don't know why they dis it that way.
Since a is a multiple of coskt, da/dt is a multiple of sinkt,
so da/dt=0 when sinkt=0 and hence v = 2 + 0 = 2 (from a)
By the result from c, at v=2 we have (why do this and not sub into a = 4cos(t/2)?)
4a^2 = 64(-(2-2)^2
a = 4
The maximum values of velocity and acceleration are 10 ms−1 and 4 ms−2 respectively.

Thanks. Sorry for the rubbish photos. I don't know how to get them to be bigger or clearer.
By the sounds of it, they've found the statationary points of acceleration, so da/dt=0. That includes both max and min. Obviously you're just interested in the max points. Is that the confusion?

Tbh, simply noting the acceleration is 8k*cos(), where k=0.5 gives the max acceleration in 1 line? Similarly for the velocity sin is max at 1, so 2+8=10.
Last edited by mqb2766; 4 weeks ago
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Thread starter 4 weeks ago
#5
(Original post by mqb2766)
By the sounds of it, they've found the statationary points of acceleration, so da/dt=0. That includes both max and min. Obviously you're just interested in the max points. Is that the confusion?

Tbh, simply noting the acceleration is 8k*cos(), where k=0.5 gives the max acceleration in 1 line? Similarly for the velocity sin is max at 1, so 2+8=10.
May be. But the book does imply that sin(3pi/2) =1. That's just wrong isn't it.
Yes, the easiest way for me to get to grips with it was to sub into the acc equation rather than using the answer to part c. I just wondered if there had been a special reason for them doing it their way. Obviously not. Thanks for clarifying. I really doubt myself when I see odd ways of doing it.
Thanks again
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4 weeks ago
#6
(Original post by maggiehodgson)
May be. But the book does imply that sin(3pi/2) =1. That's just wrong isn't it.
Yes, the easiest way for me to get to grips with it was to sub into the acc equation rather than using the answer to part c. I just wondered if there had been a special reason for them doing it their way. Obviously not. Thanks for clarifying. I really doubt myself when I see odd ways of doing it.
Thanks again
Its hard to really comment on the solution without seeing it properly and the chapter its located in (tbh Im not sure I want to). Having said that, maximsing 2+8sin() or 4cos() is a trivial thing they expect you to do by inspection (at a-level) without using calculus. Their method is tedious for this question. sin(3pi/2) is obviously -1, so its a minimum (stationary point) rather than a maximum. Its a wonder they didn't differentiate again to evaluate the sign ....

If you wanted ot use c) max acceleration is when (v-2)^2 is minimized (0) so a=4. Again, not hard.
Last edited by mqb2766; 4 weeks ago
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Thread starter 4 weeks ago
#7
(Original post by mqb2766)
Its hard to really comment on the solution without seeing it properly and the chapter its located in (tbh Im not sure I want to). Having said that, maximsing 2+8sin() or 4cos() is a trivial thing they expect you to do by inspection (at a-level) without using calculus. Their method is tedious for this question. sin(3pi/2) is obviously -1, so its a minimum (stationary point) rather than a maximum. Its a wonder they didn't differentiate again to evaluate the sign ....

If you wanted ot use c) max acceleration is when (v-2)^2 is minimized (0) so a=4. Again, not hard.
You might have the book. It's Pearson's Statistics and Mechanics Year 2 Exercise 8C q 9.
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4 weeks ago
#8
(Original post by maggiehodgson)
You might have the book. It's Pearson's Statistics and Mechanics Year 2 Exercise 8C q 9.
No, quite happy to say that :-).
Tbh, the answer reads like it was written by someone gong though the motions ... max ... have to differentiate ... rather than thinking about the question.
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4 weeks ago
#9
(Original post by maggiehodgson)
May be. But the book does imply that sin(3pi/2) =1. That's just wrong isn't it.
Yes, the easiest way for me to get to grips with it was to sub into the acc equation rather than using the answer to part c. I just wondered if there had been a special reason for them doing it their way. Obviously not. Thanks for clarifying. I really doubt myself when I see odd ways of doing it.
Thanks again
I've looked at the solution bank. It seems to me that in part (d) they've tied themselves up in knots (and made mistakes into the bargain) by trying to use the results from previous parts of the question. But the simpler route is simply to note that the maximum values of sin(kt) and cos(kt) are both 1. This give the maximum velocity and acceleration directly (and is also about right for a two point part-question).
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4 weeks ago
#10
For info:

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4 weeks ago
#11
(Original post by ghostwalker)
For info:
How can you expect students to understand proofs when the people making the solutions can't get the direction of implication right? {Sigh}
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4 weeks ago
#12
(Original post by DFranklin)
How can you expect students to understand proofs when the people making the solutions can't get the direction of implication right? {Sigh}
I hadn't actually read the question/solution.

Oh dear, that is appalling.
Last edited by ghostwalker; 4 weeks ago
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