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M2 Kinematics

When a vector is parallel to the (i-j) vector I tried to put the i part minus the j part and then equate everything to 0. This results in a wrong answer and I don't understand why..
Like looking at the ms/solution it equates the i part to the NEGATIVE j part. Why is that?
Reply 1
Original post by RK1998
When a vector is parallel to the (i-j) vector I tried to put the i part minus the j part and then equate everything to 0. This results in a wrong answer and I don't understand why..
Like looking at the ms/solution it equates the i part to the NEGATIVE j part. Why is that?


post a question and someone would help you
Reply 2
Original post by TeeEm
post a question and someone would help you


a= (5t-3)i + (8-t)j ms**

when t=0, v=(2i-5j) ms*

a) velocity after t seconds is (2.5t^2 -3t +2)i + (8t-0.5t^2 -5)j

b) Find the value of t for which is moving parallel to i-j

c) Find the speed when it is moving parallel to i-j


# I need help with part b please #
Reply 3
Original post by RK1998
a= (5t-3)i + (8-t)j ms**

when t=0, v=(2i-5j) ms*

a) velocity after t seconds is (2.5t^2 -3t +2)i + (8t-0.5t^2 -5)j

b) Find the value of t for which is moving parallel to i-j

c) Find the speed when it is moving parallel to i-j


# I need help with part b please #


When it's moving parallel to i - j, the general velocity vector will be equal to some constant multiplied by the vector I - j.
Reply 4
Original post by RK1998
Yeah but does this mean that I equate to 0 the velocity of i+j or i-j ?


When P moving parallel to
Unparseable latex formula:

\[\begin{pmatrix} \ 1, -1\end{pmatrix}\]

it means that
Unparseable latex formula:

v = \lambda \[\begin{pmatrix} \ 1, -1\end{pmatrix}\]

.
(edited 8 years ago)

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