# AS maths mechanics help?

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A particle P travels in a straight line.

At time t s, the displacement of P from a point O on the line is S metres. At time t s, the acceleration of P is (12t-4)ms-1. When t =1, s =2 and when t =3, s =30.

Find the displacement when t=2

This question is worth 8 MARKS!!!

I thought there could be some simultaneous equations involved, but I'm not really sure how to start?

Any kind of help would be great!

At time t s, the displacement of P from a point O on the line is S metres. At time t s, the acceleration of P is (12t-4)ms-1. When t =1, s =2 and when t =3, s =30.

Find the displacement when t=2

This question is worth 8 MARKS!!!

I thought there could be some simultaneous equations involved, but I'm not really sure how to start?

Any kind of help would be great!

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#2

(Original post by

A particle P travels in a straight line.

At time t s, the displacement of P from a point O on the line is S metres. At time t s, the acceleration of P is (12t-4)ms-1. When t =1, s =2 and when t =3, s =30.

Find the displacement when t=2

This question is worth 8 MARKS!!!

I thought there could be some simultaneous equations involved, but I'm not really sure how to start?

Any kind of help would be great!

**Sakura-Sama**)A particle P travels in a straight line.

At time t s, the displacement of P from a point O on the line is S metres. At time t s, the acceleration of P is (12t-4)ms-1. When t =1, s =2 and when t =3, s =30.

Find the displacement when t=2

This question is worth 8 MARKS!!!

I thought there could be some simultaneous equations involved, but I'm not really sure how to start?

Any kind of help would be great!

Now, as I am sure you know, when integrating indefinitely you get a constant of integration. Integrating indefinitely twice will then give you two constants of integration.

How lucky that we have two constraints given to us...

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(Original post by

I think the idea behind this one is to integrate twice. First to find the velocity and then integrating again to find the displacement.

Now, as I am sure you know, when integrating you get a constant of integration. Integrating twice will then give you two constants of integration.

How lucky that we have two constraints given to us...

**bluenotewitt**)I think the idea behind this one is to integrate twice. First to find the velocity and then integrating again to find the displacement.

Now, as I am sure you know, when integrating you get a constant of integration. Integrating twice will then give you two constants of integration.

How lucky that we have two constraints given to us...

thanks for the help

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#4

(Original post by

then would you integrate 12t - 4? or would you insert any values into a SUVAT equation???

thanks for the help

**Sakura-Sama**)then would you integrate 12t - 4? or would you insert any values into a SUVAT equation???

thanks for the help

Yes, I would definitely integrate. Your intuition, where the simult. eq. are concerned, is correct, I believe.

You've got this one .

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#5

I'm stuck on the exact same question I've integrated twice I've substituted the values given but I get the wrong result

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#6

(Original post by

I'm stuck on the exact same question I've integrated twice I've substituted the values given but I get the wrong result

**Bubbledubble**)I'm stuck on the exact same question I've integrated twice I've substituted the values given but I get the wrong result

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#9

**Sakura-Sama**)

A particle P travels in a straight line.

At time t s, the displacement of P from a point O on the line is S metres. At time t s, the acceleration of P is (12t-4)ms-1. When t =1, s =2 and when t =3, s =30.

Find the displacement when t=2

This question is worth 8 MARKS!!!

I thought there could be some simultaneous equations involved, but I'm not really sure how to start?

Any kind of help would be great!

You integrate twice to come up with an equation for displacement and end up with two values. They've given you two sets of data that you can substitute in and solve simultaneously. With your values of the two variables, plug them into the equation for s and you'll end up with an answer.

This may be wrong, but I got s = 6m as my answer.

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#11

(Original post by

Not sure how late I am since I'm here looking for an answer so I'm assuming others are too, but I experimented and ended up using simultaneous equations.

You integrate twice to come up with an equation for displacement and end up with two values. They've given you two sets of data that you can substitute in and solve simultaneously. With your values of the two variables, plug them into the equation for s and you'll end up with an answer.

This may be wrong, but I got s = 6m as my answer.

**kanvi06**)Not sure how late I am since I'm here looking for an answer so I'm assuming others are too, but I experimented and ended up using simultaneous equations.

You integrate twice to come up with an equation for displacement and end up with two values. They've given you two sets of data that you can substitute in and solve simultaneously. With your values of the two variables, plug them into the equation for s and you'll end up with an answer.

This may be wrong, but I got s = 6m as my answer.

(Original post by

This like really helped me, and I got 6m as the answer as well

**L.Da**)This like really helped me, and I got 6m as the answer as well

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#12

I'm still confused about the simultaneous equations parts ? I don't know if I got the equation for s right

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#13

(Original post by

I'm still confused about the simultaneous equations parts ? I don't know if I got the equation for s right

**anonymous1406**)I'm still confused about the simultaneous equations parts ? I don't know if I got the equation for s right

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#15

(Original post by

my first integration I got v=6t^2-4t and I integrated that to get s=2t^3-2t^2

**anonymous1406**)my first integration I got v=6t^2-4t and I integrated that to get s=2t^3-2t^2

So, you'd have . And then repeat the process to get s, adding a second constant of integration (call it "d").

Then you can look to use simultaneous equations to find the two contants and hence the formula for s.

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#16

**anonymous1406**)

my first integration I got v=6t^2-4t and I integrated that to get s=2t^3-2t^2

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#17

I knew that was what I was missing I wrote plus c but couldn't think how to find it having a mind blank

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#20

https://imgur.com/a/AtT31AAMy writing isn't the best, but it should be readable (I have skipped over solving the simultaneous equations as they are not difficult)

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