# M2Watch

Announcements
#1
I was having some trouble with M2 edexcel

so the question is:

https://6caf30cbcaf0c74dfdacf3d63c55...EZJY1U/CH1.pdf

It is exercise 1C question 12

I have gotten the first part so that r=45i+67.5j when t=3

at t seconds A=10i+15j

and that v=5t^2i+15tj

now for the second part of the question I worked out a new acceleration since the force had changed

so now a=12i+ (60-t^2/0.2)j

And now I don't know what to do and was wondering if someone could help guide me?
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4 years ago
#2
Your velocity in the vertical component will have the speed of 0 when the particle is at it's greatest height.
Quote me if you still don't get it.
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#3
(Original post by Placeboo123)
Your velocity in the vertical component will have the speed of 0 when the particle is at it's greatest height.
Quote me if you still don't get it.
Hi sorry but I don't understand I am looking at exercise 1C question 12 where does greatest height fit in?
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4 years ago
#4
(Original post by Teddysmith123)
Hi sorry but I don't understand I am looking at exercise 1C question 12 where does greatest height fit in?
Oh sorry I didn't see the other questions.
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#5
(Original post by Placeboo123)
Oh sorry I didn't see the other questions.

It's ok do you think you could help me with question 12 exercise 1c?
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4 years ago
#6
(Original post by Teddysmith123)
It's ok do you think you could help me with question 12 exercise 1c?
Sorry my laptop died.
Ehm I've got the same answer as you for the first part.
For the second part first of all note down your velocity at t = 3 using what you've already found in part 1.
Then you'll have to find new velocity by integrating the acceleration which you'll get from F=MA and this time you'll have two constants from part 1.
This is what springs into mind when I read the question, I'll answer it too tell you what I get.
Hope you know what to do now.
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#7
(Original post by Placeboo123)
Sorry my laptop died.
Ehm I've got the same answer as you for the first part.
For the second part first of all note down your velocity at t = 3 using what you've already found in part 1.
Then you'll have to find new velocity by integrating the acceleration which you'll get from F=MA and this time you'll have two constants from part 1.
This is what springs into mind when I read the question, I'll answer it too tell you what I get.
Hope you know what to do now.

See I subbed t=3 into my velocity from part 1 then I also found a new velocity from the new acceleration in 2. I then let t=3 in the new velocity equation and equated v from part 1 with v from part 2 and got a value for c. But then when I tried to sub in t=6 my answer was different to the one that was in the book!
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4 years ago
#8
(Original post by Teddysmith123)
See I subbed t=3 into my velocity from part 1 then I also found a new velocity from the new acceleration in 2. I then let t=3 in the new velocity equation and equated v from part 1 with v from part 2 and got a value for c. But then when I tried to sub in t=6 my answer was different to the one that was in the book!
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#9
(Original post by Placeboo123)

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4 years ago
#10
(Original post by Teddysmith123)
That's what I've got.
I think where you went wrong is you've taken t=6 instead of 3
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#11
(Original post by Placeboo123)
That's what I've got.
I think where you went wrong is you've taken t=6 instead of 3

If I write my working out can you see where I have gone wrong:

so first I didvide the new force by 0.2 to get the new acceleration which is 30i+(60/5t^2)

then I integrate this to get V= 30ti+(60t-5/3 t^3)j +c

I then let t=3 for V and equate this with t=3 in the V i found in part 1

I then get a c value of 45i+90j

I then sub c into v in part 2 and then let t=6
0
4 years ago
#12
(Original post by Teddysmith123)
If I write my working out can you see where I have gone wrong:

so first I didvide the new force by 0.2 to get the new acceleration which is 30i+(60/5t^2)

then I integrate this to get V= 30ti+(60t-5/3 t^3)j +c

I then let t=3 for V and equate this with t=3 in the V i found in part 1

I then get a c value of 45i+90j

I then sub c into v in part 2 and then let t=6
Check your velocity at t = 3 again.
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#13
(Original post by Placeboo123)
Check your velocity at t = 3 again.

I did and I keep getting the same answer! Can you post your working so that I can see where I have gone wrong?
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4 years ago
#14
(Original post by Teddysmith123)
I did and I keep getting the same answer! Can you post your working so that I can see where I have gone wrong?
WOw this is incredible I've misread the question to say (2ti - 3j) in the first part instead of (2ti + 3j) so I've got 45i - 45j as velocity at t=3, v = 5t^2i - 15tj.
It all fell into place with my mistake.
Now I'm wondering too.
Btw
(Original post by Teddysmith123)
I then get a c value of 45i+90j
(Original post by Teddysmith123)
and that v=5t^2i+15tj
15*3 is 45 not 90.

When I accidently took it to be negative it worked perfectly, how reliable is the answer you've got? Maybe they've done the same mistake as me.
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#15
(Original post by Placeboo123)
WOw this is incredible I've misread the question to say (2ti - 3j) in the first part instead of (2ti + 3j) so I've got 45i - 45j as velocity at t=3, v = 5t^2i - 15tj.
It all fell into place with my mistake.
Now I'm wondering too.
Btw

15*3 is 45 not 90.

When I accidently took it to be negative it worked perfectly, how reliable is the answer you've got? Maybe they've done the same mistake as me.

I'm not entierly sure how reliable the book is since I checked a few other tsr forums for M2 and apparently the book is riddled with mistakes but yes when I take a negative value I get the answer they are looking for but when I don't I get the incorrect answer and it doesn't make any sense to me!
0
4 years ago
#16
(Original post by Teddysmith123)
I'm not entierly sure how reliable the book is since I checked a few other tsr forums for M2 and apparently the book is riddled with mistakes but yes when I take a negative value I get the answer they are looking for but when I don't I get the incorrect answer and it doesn't make any sense to me!
Another thing that makes me think is when we take for the j component t = 6.
v = 60t-5/3 t^3 + 45 we actually get 45 too!
0
4 years ago
#17
(Original post by Teddysmith123)
I'm not entierly sure how reliable the book is since I checked a few other tsr forums for M2 and apparently the book is riddled with mistakes but yes when I take a negative value I get the answer they are looking for but when I don't I get the incorrect answer and it doesn't make any sense to me!
Hahaha what the hell, if the 2i component was negative in the first part of the question and we'd use t=6 you'd also get their answer.

Though I think we'd done it correctly.
Any more questions? that was interesting, did some revision for once! haha
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#18
(Original post by Placeboo123)
Hahaha what the hell, if the 2i component was negative in the first part of the question and we'd use t=6 you'd also get their answer.

Though I think we'd done it correctly.
Any more questions? that was interesting, did some revision for once! haha
Haha no this was the only question in the exercise that had me stumped! Are you also doing further maths? I just have m2 and s2 left to complete but so far m2 seems like a pain...
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4 years ago
#19
(Original post by Teddysmith123)
Haha no this was the only question in the exercise that had me stumped! Are you also doing further maths? I just have m2 and s2 left to complete but so far m2 seems like a pain...
Yeah doing 8 maths modules this year.
I saw your post about FP3, what's that like compared to FP1 or anything I'll be doing that next year
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#20
(Original post by Placeboo123)
Yeah doing 8 maths modules this year.
I saw your post about FP3, what's that like compared to FP1 or anything I'll be doing that next year
Fp3 is ok I prefer it to fp2. The vectors are a bit fiddly but not too bad. Wow 8 modules thats intense! I have 6 modules this yeat to complete the whole a level in fm. Well thanks for helping me with my question and good luck...

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