AS Mechanics Question

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Anon134132523
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#1
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#1
A particle of mass, M, which is subject to a constant resisting force, is observed to cover distances of s1, and sn respectively in the first and nth seconds of its motion. Find an expression in terms of s1 and sn for the magnitude of the resistance force. How do I make this equation?
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ghostwalker
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#2
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(Original post by Anon134132523)
A particle of mass, M, which is subject to a constant resisting force, is observed to cover distances of s1, and sn respectively in the first and nth seconds of its motion. Find an expression in terms of s1 and sn for the magnitude of the resistance force. How do I make this equation?
Resistance is constant, so acceleration/deceleration is constant, and hence suvat applies.

Consider how far does it travel in the first second? And the n'th second?

You now have two simultaneous equations....
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Anon134132523
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#3
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So far I've got:
Part one:
s = s1, u = ?, v = ?, a = a, t= 1
(s + 1/2at^2) / t = v
v = s1 + a/2
Part two:
s = sn- s1, u = s1 + a/2, v = ?, a = a, t = n-1
(s - 1/2at^2)/t = u
((sn-s1) - a/2(n-1)^2) / (n-1) = u
As the final velocity of part one is the initial velocity of part two:
((sn-s1) - a/2(n-1)^2) / (n-1) = s1 + a/2
However, I can't seem to get the right answer from there?
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ghostwalker
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(Original post by Anon134132523)
So far I've got:
Part one:
s = s1, u = ?, v = ?, a = a, t= 1
(s + 1/2at^2) / t = v
v = s1 + a/2
Part two:
s = sn- s1, u = s1 + a/2, v = ?, a = a, t = n-1
(s - 1/2at^2)/t = u
((sn-s1) - a/2(n-1)^2) / (n-1) = u
As the final velocity of part one is the initial velocity of part two:
((sn-s1) - a/2(n-1)^2) / (n-1) = s1 + a/2
However, I can't seem to get the right answer from there?
"As the final velocity of part one is the initial velocity of part two:"

This is untrue. The final velocity for the first part will be the velocity at time t=1. This is not the same as the velocity at the start of the second interval. So, you can't use "v" to tie the two together.

Do you know what the answer is supposed to be?

The distance travelled in the first second is simply s1= u + a/2

What about the n'th second, the interval between t=n-1 and t=n (big hint)
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Anon134132523
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#5
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(Original post by ghostwalker)
"As the final velocity of part one is the initial velocity of part two:"

This is untrue. The final velocity for the first part will be the velocity at time t=1. This is not the same as the velocity at the start of the second interval. So, you can't use "v" to tie the two together.

Do you know what the answer is supposed to be?

The distance travelled in the first second is simply s1= u + a/2

What about the n'th second, the interval between t=n-1 and t=n (big hint)
Sn = u + 1/2a(2n-1)
u = S1 - a/2
Sn = (sn - a/2) + 1/2a(2n-1)
Sn - S1 = -1/2a(1-(2n-1))
(-2Sn+2S1) / (-2n+2) = a
a = (Sn-S1)/(n-1)?
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ghostwalker
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#6
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(Original post by Anon134132523)
Sn = u + 1/2a(2n-1)
u = S1 - a/2
Sn = (sn - a/2) + 1/2a(2n-1)
Sn - S1 = -1/2a(1-(2n-1))
(-2Sn+2S1) / (-2n+2) = a
a = (Sn-S1)/(n-1)?
That's what I got.
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Anon134132523
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#7
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(Original post by ghostwalker)
That's what I got.
Thanks a lot man
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ghostwalker
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#8
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(Original post by Anon134132523)
Thanks a lot man
np.

Don't forget the question asked for the force, not the accleration.
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