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M1 Vector question help

Need some help with this question:

The velocity, v ms-1, of a particle P at time t seconds is given by

v = (1-2t)i​ + (3t-3)j

Find the value of t when P is moving parallel to (-i - 3j)
Reply 1
Avatar for Lulu24
Lulu24
13
Original post by awkwardesiguy
Need some help with this question:

The velocity, v ms-1, of a particle P at time t seconds is given by

v = (1-2t)i​ + (3t-3)j

Find the value of t when P is moving parallel to (-i - 3j)


hmmm, so the particle is moving at (1, -2t ) + (3t - 3), so you need to find t when (1-2t ) = -1 and (3t-3 ) = 3, right?
Original post by Lulu24
hmmm, so the particle is moving at (1, -2t ) + (3t - 3), so you need to find t when (1-2t ) = -1 and (3t-3 ) = 3, right?


I tried doing that, but what I did was:

(1 - 2t) = -1

Hence, -3(1 - 2t) = 3

Therefore, -3(1 - 2t) = (3t - 3)

Which is giving me no value for t
Reply 3
Avatar for M14B
M14B
18
Original post by awkwardesiguy
i tried doing that, but what i did was:

(1 - 2t) = -1

hence, -3(1 - 2t) = 3

therefore, -3(1 - 2t) = (3t - 3)

which is giving me no value for t



post it in maths!
Original post by M14B
post it in maths!


I shall, thank you for your help though!
Original post by awkwardesiguy
Need some help with this question:

The velocity, v ms-1, of a particle P at time t seconds is given by

v = (1-2t)i​ + (3t-3)j

Find the value of t when P is moving parallel to (-i - 3j)


Since they are parallel their angles wit the horizontal(tan) should be same,so if you equate their angles you will get the answer.And the operation you did is completely against the rules of algebra:wink:..you cannot equate values for the number that is you cannot substitute 3t-3 for 3.

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