Mechanics Problem with I/J vectors

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The discussion revolves around the motion of two particles, P and Q, in an x-y plane, defined by their position vectors: P as (t²†i + 4t†j)m and Q as (2t†i + (t + 1)†j)m. It is established that the particles never collide, as their position vectors do not equal each other at the same time. The velocity of particle Q is constant, calculated as 2†i + 1†j with a magnitude of √5 m/s and a direction of 26.57°. For parallel velocities, the correct time t is determined to be 4, not 1, by ensuring the velocity components are proportional rather than equal.

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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 calculate 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


?

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 :)
 
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FaraDazed said:

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 calculate 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


?


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 :)
Hello FaraDazed,
For part C you are asked to find t for parallel vectors.What you are doing is making them equal (by equating both components ).When two vectors are parallel their components are proportional. I think this should give you the correct answer.
If you use t=1 as you got in C
v(p)=(2\hat{i}+4\hat{j})
and v(q)=(2\hat{i}+\hat{j})
They are neither equal nor parallel.
Just try making the velocity components proportional.
Regards
Yukoel
 
If two vectors are parallel, then one can be expressed as a scalar multiple of the other. i.e we can express this as $$\begin{pmatrix} {2}\\{1} \end{pmatrix} = \alpha \begin{pmatrix} {2t}\\{4} \end{pmatrix}$$ Solve this system for ##\alpha## and hence for ##t##.
 
Thank you guys I get it now, thanks :)
 

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