# AS maths mechanics help?

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#1
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! 0
2 years ago
#2
(Original post by 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! 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 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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#3
(Original post by 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...
then would you integrate 12t - 4? or would you insert any values into a SUVAT equation???
thanks for the help 0
2 years ago
#4
(Original post by Sakura-Sama)
then would you integrate 12t - 4? or would you insert any values into a SUVAT equation???
thanks for the help I don't think SUVAT applies here. We are not told (in fact the acceleration is clearly non-constant as it depends on t) that the acceleration is constant, which is a necessary condition for suvat to work.
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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1 year ago
#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 0
1 year ago
#6
(Original post by 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 0
1 year ago
#7
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5 months ago
#8
Did you find out what the answer was?
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4 months ago
#9
(Original post by 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! 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.
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4 months ago
#10
This like really helped me, and I got 6m as the answer as well 0
4 months ago
#11
(Original post by 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 L.Da)
This like really helped me, and I got 6m as the answer as well This thread is 2 years old but I can confirm that the answer is 6m.
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4 months ago
#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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4 months ago
#13
(Original post by anonymous1406)
I'm still confused about the simultaneous equations parts ? I don't know if I got the equation for s right
I think I've chucked my workings now - why don't you upload what you get when you integrate and we can check it!
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4 months ago
#14
my first integration I got v=6t^2-4t and I integrated that to get s=2t^3-2t^2
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4 months ago
#15
(Original post by anonymous1406)
my first integration I got v=6t^2-4t and I integrated that to get s=2t^3-2t^2
You've missed out the constants of integration (2 of them).

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.
2
4 months ago
#16
(Original post by anonymous1406)
my first integration I got v=6t^2-4t and I integrated that to get s=2t^3-2t^2
Don't forget your arbitrary constants - you'll need one after the 1st integration, then that integrates to something else with another constant added on!
1
4 months ago
#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
0
4 months ago
#18
If anyone's still confused then I can upload a picture of my working? Would that help?
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4 months ago
#19
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4 months ago
#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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