FaraDazed
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Homework Statement
Two particles P and Q move in an x-y plane. At time t seconds the position vector of P is (t^2\hat{i}+4t\hat{j})m and Q is (2t\hat{i}+(t+1)\hat{j})m.
A: Prove the particle never collide
B: Show that the velocity of Q is constant, and calcualte the magnitude and direction
C: Find the value for t when P and Q have parallel velocities and find the distance between them at this point.
Homework Equations
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The Attempt at a Solution
Its part C that is confusing me but incase I messed the others up I will post them as well.
Part A:
For the particles to collide their position vectors will equal each other at the same time so setting their I an J components to equal each other should produce any times they are equal. First their I components.
<br /> 2t=t^2 \\<br /> 0=t^2 - 2t \\<br /> 0=t(t-2)<br />
Therefore when t=0 and t=2 the I components are the same. And now the J components.
<br /> t+1=4t \\<br /> 1=3t \\<br /> t=\frac{1}{3}<br />
Therefore their J components are only equal when t=1/3, therefore they never collide as the I and J's are never the same at the same time.
Part B:
The velocity of Q will be constant if neither of the components are functions of time so differentiation will find me the velocity vector
<br /> \dot{r_Q}=2\hat{i}+1\hat{j}<br />
Therefore the velocity is constant as neither are functions of time. To find the magnitude next
<br /> \sqrt{2^2+1^2}=\sqrt{5}m/s<br />
And the direction
<br /> tan^{-1}(\frac{1}{2})=26.57°<br />
Part C:
To find the time when P and Q have parallel velocities first P's velocity vector needs to be found
<br /> \dot{r_P}=2t\hat{i}+4\hat{j}<br />
It can be seen that both P and Q have constant J components therefore they will be parallel when the I components are the same.
<br /> 2t=2 \\<br /> t=1<br />
However the answer to part C is apparently 4 not 1. Any help is appreciated :)