# Help with M1 question please?

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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..

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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(Original post by

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..

**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..

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#3

you said you found out the acceleration a which is such that

.

integrate once wrt t

integrate wrt t again

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

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

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.

.

integrate once wrt t

**(*)**, E an integration constant. This also gives you the velocity of the cyclist.integrate wrt t again

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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(Original post by

Use 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.

**Gemini92**)Use 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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#6

(Original post by

I did that and got 19.6 :s

**Sparkzz**)I did that and got 19.6 :s

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(Original post by

That's strange, I got 15. What did you get as the acceleration?

**Gemini92**)That's strange, I got 15. What did you get as the acceleration?

as the equation writes

144=64+40a

80=40a

a = 2?

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#8

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(Original post by

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.

**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.

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#11

(Original post by

i got a =1 because it is 2 * a * 40 then solve it go find that a is 1

**F1 GUJ**)i got a =1 because it is 2 * a * 40 then solve it go find that a is 1

**7-year-old**thread?

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