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Mechanics Problem with I/J vectors

  • Thread starter FaraDazed
  • Start date
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
?


3. 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 :)
 
73
1
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
?


3. 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 :)
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)=[itex](2\hat{i}+4\hat{j})[/itex]
and v(q)=[itex](2\hat{i}+\hat{j})[/itex]
They are neither equal nor parallel.
Just try making the velocity components proportional.
Regards
Yukoel
 

CAF123

Gold Member
2,882
87
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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