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M1 Vector Integral

My answer is different to that of mark scheme. I am probably being thick but I just can't spot my error. Couldn't upload photos so heres a link to them
https://plus.google.com/110645167952183816519/posts/GobAfUyf9iG
Reply 1
Avatar for Zacken
Zacken
22
Original post by Nels98
My answer is different to that of mark scheme. I am probably being thick but I just can't spot my error. Couldn't upload photos so heres a link to them
https://plus.google.com/110645167952183816519/posts/GobAfUyf9iG


Your integration is wrong: v=∫0i−2j dt=ci−(2t+d)j\mathbf{v} =\int 0i - 2j \, \mathrm{d}t = ci - (2t + d)j.

Then, 4i+0j=ci−dj⇒c=4,d=04i + 0j = ci - dj \Rightarrow c =4, d=0.
Reply 2
Avatar for Nels98
Nels98
OP
2
Original post by Zacken
Your integration is wrong: v=∫0i−2j dt=ci−(2t+d)j\mathbf{v} =\int 0i - 2j \, \mathrm{d}t = ci - (2t + d)j.

Then, 4i+0j=ci−dj⇒c=4,d=04i + 0j = ci - dj \Rightarrow c =4, d=0.


thats no different isn't it? surely that still ends up with v=4i-2tj
Reply 3
Avatar for Nels98
Nels98
OP
2
(i think) The only issue is that they have a t in the position vector for the j term whereas i have a t^2.
Reply 4
Avatar for Zacken
Zacken
22
Original post by Nels98
thats no different isn't it? surely that still ends up with v=4i-2tj


Ah, yes - fair enough. I misread. I agree with your final answer.
Reply 5
Avatar for Nels98
Nels98
OP
2
Original post by Zacken
Ah, yes - fair enough. I misread. I agree with your final answer.


hmm seems strange for an OCR MEI mark scheme to be wrong but thanks for confirming my answer
Reply 6
Avatar for Zacken
Zacken
22
Original post by Nels98
hmm seems strange for an OCR MEI mark scheme to be wrong but thanks for confirming my answer


It's probably more a typographical error on their part, oh well.
Agree with OP and Zacken. t^2
Reply 8
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Nels98
OP
2
ok good stuff, thanks!

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