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1. Just wondering (something like this was on my exam)

if you have a position vector: (making this up)

r = (2t-0.04t^2)i + (3t+0.06t^2)j

and the question says find the time that the particle is heading south east..how would go about doing it?

thanks
2. South east means the angle it makes with the +ve i coord is 45. So...

tan45 = -j/i = -(3t+0.06t^2) / (2t-0.04t^2)

2t-0.04t^2 = -3t-0.06t^2 [since tan45=1]
3. When it's heading south east, it will be travelling in the direction k(i - j) where k is a scalar so you put 2t-0.04t^2 = -(3t+0.06t^2) and solve
4. right..nice one bezza and shift3, i think i got that right then..phew.

thanks.
5. (Original post by ShOcKzZ)
Just wondering (something like this was on my exam)

if you have a position vector: (making this up)

r = (2t-0.04t^2)i + (3t+0.06t^2)j

and the question says find the time that the particle is heading south east..how would go about doing it?

thanks
At this time i = -j, so:
2t - 0.04t^2 = -(3t + 0.06t^2)
2t - 0.04t^2 = -3t - 0.06t^2
5t + 0.02t^2 = 0
t(5 + 0.02t) = 0
5 = -0.02t
t = -250 s

Huh? Should this not be positive?
6. (Original post by mik1a)
At this time i = -j, so:
2t - 0.04t^2 = -(3t + 0.06t^2)
2t - 0.04t^2 = -3t - 0.06t^2
5t + 0.02t^2 = 0
t(5 + 0.02t) = 0
5 = -0.02t
t = -250 s

Huh? Should this not be positive?
t(5+0.02t) = 0
t = 0 or t = -250, in which case you'd take t=0.
7. (Original post by ShOcKzZ)
Just wondering (something like this was on my exam)

if you have a position vector: (making this up)

r = (2t-0.04t^2)i + (3t+0.06t^2)j

and the question says find the time that the particle is heading south east..how would go about doing it?

thanks
Wait a sec, when it is travelling south east you need to differentiate wrt. time:

r = (2t-0.04t^2)i + (3t+0.06t^2)j
dr/dt = (2 - 0.08t)i + (3 + 0.12t)j

Velocity south east = k(i-j), so i = -j
(2 - 0.08t) = -(3 + 0.12t)
2 - 0.08t = -3 -0.12t
5 + 0.04t = 0
t = -125 s

What you all found (and me admittedly) was when the particle was on the line (i-j) from the origin, but it was probably travelling in a different direction!
8. (Original post by mik1a)
What you all found (and me admittedly) was when the particle was on the line (i-j) from the origin, but it was probably travelling in a different direction!

Good point
9. (Original post by ShOcKzZ)
Just wondering (something like this was on my exam)

if you have a position vector: (making this up)

r = (2t-0.04t^2)i + (3t+0.06t^2)j

and the question says find the time that the particle is heading south east..how would go about doing it?

thanks

i = -j
2t - 0.04t^2 = -(3t + 0.06t^2)
2t - 0.04t^2 = -0.06t^2 - 3t
0.02t^2 + 5t = 0
t(0.02t + 5) = 0

t = 0 seconds OR:

0.02t = -5
t = -5/(0.02)
t = -250 seconds

But as time can't have a -ve value:

t = 0 seconds

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