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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 [itex](t^2\hat{i}+4t\hat{j})m[/itex] and Q is [itex](2t\hat{i}+(t+1)\hat{j})m[/itex].

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 paralell 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 eachother should produce any times they are equal. First their I components.

[tex]

2t=t^2 \\

0=t^2 - 2t \\

0=t(t-2)

[/tex]

Therefore when t=0 and t=2 the I components are the same. And now the J components.

[tex]

t+1=4t \\

1=3t \\

t=\frac{1}{3}

[/tex]

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

[tex]

\dot{r_Q}=2\hat{i}+1\hat{j}

[/tex]

Therefore the velocity is constant as neither are functions of time. To find the magnitude next

[tex]

\sqrt{2^2+1^2}=\sqrt{5}m/s

[/tex]

And the direction

[tex]

tan^{-1}(\frac{1}{2})=26.57°

[/tex]

Part C:

To find the time when P and Q have parallel velocities first P's velocity vector needs to be found

[tex]

\dot{r_P}=2t\hat{i}+4\hat{j}

[/tex]

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.

[tex]

2t=2 \\

t=1

[/tex]

However the answer to part C is apparently 4 not 1. Any help is appreciated :)