Mechanics Problem with I/J vectors

[FaraDazed]

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.
$$2t=t^2 \\ 0=t^2 - 2t \\ 0=t(t-2)$$
Therefore when t=0 and t=2 the I components are the same. And now the J components.
$$t+1=4t \\ 1=3t \\ t=\frac{1}{3}$$
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
$$\dot{r_Q}=2\hat{i}+1\hat{j}$$
Therefore the velocity is constant as neither are functions of time. To find the magnitude next
$$\sqrt{2^2+1^2}=\sqrt{5}m/s$$
And the direction
$$tan^{-1}(\frac{1}{2})=26.57°$$

Part C:
To find the time when P and Q have parallel velocities first P's velocity vector needs to be found
$$\dot{r_P}=2t\hat{i}+4\hat{j}$$
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.
$$2t=2 \\ t=1$$

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

 

[Yukoel]

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

?

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.
$$2t=t^2 \\ 0=t^2 - 2t \\ 0=t(t-2)$$
Therefore when t=0 and t=2 the I components are the same. And now the J components.
$$t+1=4t \\ 1=3t \\ t=\frac{1}{3}$$
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
$$\dot{r_Q}=2\hat{i}+1\hat{j}$$
Therefore the velocity is constant as neither are functions of time. To find the magnitude next
$$\sqrt{2^2+1^2}=\sqrt{5}m/s$$
And the direction
$$tan^{-1}(\frac{1}{2})=26.57°$$

Part C:
To find the time when P and Q have parallel velocities first P's velocity vector needs to be found
$$\dot{r_P}=2t\hat{i}+4\hat{j}$$
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.
$$2t=2 \\ t=1$$

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

 

[CAF123]

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 $t$.

 

[FaraDazed]

Thank you guys I get it now, thanks :)

 

[Nickie]

For part C, you have correctly found that the parallel velocities occur when t=1. However, the distance between P and Q at this time is not 1, but rather:

r_P = (1^2*1 + 4*1) = (1+4) = 5m
r_Q = (2*1 + (1+1)) = (2+2) = 4m

The distance between P and Q at this time is the magnitude of the difference between their position vectors:

|r_P-r_Q| = |(5-4)\hat{i} + (4-4)\hat{j}| = |1\hat{i} + 0\hat{j}| = sqrt(1^2 + 0^2) = 1m

Therefore, the distance between P and Q at the time when they have parallel velocities is 1m.