Sparkzz
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A cyclist travelling with constant acceleration along a straight road, passes three points, A, B and C where AB=BC=20m. The speed of the cyclist at A is 8ms-1 and at B is 12ms-1. Find the speed of the cyclist at C.

I found out the acceleration and then cyclist making the displacement 40m??

Answer is 15 though, dno how to arrive at that..
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Gemini92
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(Original post by Sparkzz)
A cyclist travelling with constant acceleration along a straight road, passes three points, A, B and C where AB=BC=20m. The speed of the cyclist at A is 8ms-1 and at B is 12ms-1. Find the speed of the cyclist at C.

I found out the acceleration and then cyclist making the displacement 40m??

Answer is 15 though, dno how to arrive at that..
Use v^2 = u^2 + 2as to find the acceleration between A and B, where v=12, u=8 and s=20. You then put this acceleration back into the same formula to calculate v between A and C, where u=8, s=40 and a=whatever you calulated it to be.
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zeratul
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you said you found out the acceleration a which is such that
\dfrac{d^2x}{dt^2}=a.
integrate once wrt t
\dfrac{dx}{dt}=at+E (*), E an integration constant. This also gives you the velocity of the cyclist.
integrate wrt t again
x=at^2/2+Et+F where F is another integration constant

Without loss of generality say that x=0 when the cyclist is at A (so F=0), where he has (initially) speed 8, so substitute v=8 at t=0 in (*) to get that E=8.

Now put x=40 when the cyclist is at C, and thus 40=at^2/2+8t, and you can determine t when x=40.
Then put the t you just determined in (*) to get the speed at C.

I have 2 comments:
1. My solution might not be the best, this is definitely not my field.
2. Maybe you're wondering why I haven't used the given speed at B. Well, you said you knew a, so I implicitly used this condition when I used the 'known' acceleration a.
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zeratul
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(Original post by Gemini92)
v^2 = u^2 + 2as
My long-winded solution blatantly avoids this very simple formula, definitely the best way to do this.
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Sparkzz
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(Original post by Gemini92)
Use v^2 = u^2 + 2as to find the acceleration between A and B, where v=12, u=8 and s=20. You then put this acceleration back into the same formula to calculate v between A and C, where u=8, s=40 and a=whatever you calulated it to be.
I did that and got 19.6 :s
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Gemini92
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(Original post by Sparkzz)
I did that and got 19.6 :s
That's strange, I got 15. What did you get as the acceleration?
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Sparkzz
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(Original post by Gemini92)
That's strange, I got 15. What did you get as the acceleration?
acceleration 2?

as the equation writes
144=64+40a
80=40a
a = 2?
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Gemini92
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(Original post by Sparkzz)
acceleration 2?

as the equation writes
144=64+40a
80=40a
a = 2?
That's what I got. If you put that back into the same formula with u=8 and s=40, you should get v=15.
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Sparkzz
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(Original post by Gemini92)
That's what I got. If you put that back into the same formula with u=8 and s=40, you should get v=15.
oh yeh woops im so stupid, did s=80 lol thank u!!
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F1 GUJ
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i got a =1 because it is 2 * a * 40 then solve it go find that a is 1
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Prasiortle2
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(Original post by F1 GUJ)
i got a =1 because it is 2 * a * 40 then solve it go find that a is 1
No, we know the speeds at B and C, and the displacement from B to C is 20 m, so it's 2\times a \times 20. Also, why did you bump a 7-year-old thread?
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