I with this question about vectors

  • #1
lioric
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Homework Statement


At noon two boats P and Q have a position vector (i+7j)km and (3i+8j)km respectively to the origin O.
i and j are unit vectors in direction to East and north respectively
P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
At time t after noon the position vectors of P and Q with respect to O are p km and Q km
a) show that the velocity of P is ( 5i + 5j)
b) find the time to the nearest minute when Q is north east of P

Homework Equations


Phythogorus theorum a^2 = b^2 + c^2


The Attempt at a Solution


I have done part a)
Where if the velocity is (5i+5j), then if you use phythogorus theorum where √(5^2+5^2) I get 5√2

I don't know how to do part b)
 

Answers and Replies

  • #2
PeroK
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Have you drawn a diagram?
 
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  • #3
LCKurtz
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Homework Statement


At noon two boats P and Q have a position vector (i+7j)km and (3i+8j)km respectively to the origin O.
i and j are unit vectors in direction to East and north respectively
P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
At time t after noon the position vectors of P and Q with respect to O are p km and Q km
a) show that the velocity of P is ( 5i + 5j)
The red quotes seem inconsistent.
 
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  • #4
phinds
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Have you drawn a diagram?
Yeah. What he said.

@lioric discussing this kind of problem without a vector diagram is a waste of time.
 
  • #5
lioric
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Yeah. What he said.

@lioric discussing this kind of problem without a vector diagram is a waste of time.
Yes I have drawn a diagram
 
  • #6
phinds
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Yes I have drawn a diagram
Uh ... and do you plan to share it with US? You ARE asking us to discuss it after all.
 
  • #7
lioric
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But my main problem is figuring out the wording of the question
Is the question asking me to find the time when Q is at a bearing of 45 degrees of P?
 
  • #8
phinds
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But my main problem is figuring out the wording of the question
Is the question asking me to find the time when Q is at a bearing of 45 degrees of P?
What does "North East" mean?
 
  • #9
lioric
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What does "North East" mean?
East is 90 degrees from North
North east is the bisection of that
It's the line equidistant from the East and the north
The angle is 45 degrees from north
 
  • #10
phinds
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East is 90 degrees from North
North east is the bisection of that
It's the line equidistant from the East and the north
The angle is 45 degrees from north
There are two lines that are 45 degrees from North. Yes, I DO know what you mean but my point is that without a vector diagram, again, this discussion is useless.
 
  • #11
LCKurtz
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@lioric: Are you ever going to respond to post #3 with a correct statement of the problem?
 
  • #13
lioric
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@lioric: Are you ever going to respond to post #3 with a correct statement of the problem?
What can I say about what the red quotes say? It's what the question said. Word to word
 
  • #14
lioric
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AND a vector diagram !
I have drawn the diagram
And I can solve the problem when I'm sure about the north east part
I just need to know the exact meaning of that
Just clearing my doubts
 
  • #15
LCKurtz
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What can I say about what the red quotes say? It's what the question said. Word to word
Do you understand that the direction and vector I quoted in red do not agree?
 
  • #16
phinds
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Do you understand that the direction and vector I quoted in red do not agree?
He might have figured that out if he had done a correct vector diagram, which is one reason I keep telling him to do one (AND show it to us so we can help)
 
  • #17
lioric
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Sorry that's a typo
It's 5i-5j
Sorry
 
  • #18
lioric
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At noon two boats P and Q have a position vector (i+7j)km and (3i-8j)km respectively to the origin O.
i and j are unit vectors in direction to East and north respectively
P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
At time t after noon the position vectors of P and Q with respect to O are p km and Q km
a) show that the velocity of P is ( 5i - 5j)
b) find the time to the nearest minute when Q is north east of P

This is the correct one
 
  • #19
lioric
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IMG_20171225_171312.jpg
 

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  • #20
LCKurtz
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Sorry that's a typo
It's 5i-5j
Sorry
That and the now changed original position of Q. So now you can show us how you calculate Q and P's positions at time t. Then think about what must be true about Q-P for Q to be northeast of P.
 
  • #21
lioric
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That and the now changed original position of Q. So now you can show us how you calculate Q and P's positions at time t. Then think about what must be true about Q-P for Q to be northeast of P.
Q must be at a 45 degrees bearing to P
 
  • #22
LCKurtz
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I posted: "So now you can show us how you calculate Q and P's positions at time t". Which you ignored.
 
  • #23
lioric
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I posted: "So now you can show us how you calculate Q and P's positions at time t". Which you ignored.
We use r=r0 + vt
 
  • #24
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We use r=r0 + vt
LCKurtz's question was more specific than this. In other words, what were your calculations for Q and P at an arbitrary time t?
 
  • #25
lioric
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LCKurtz's question was more specific than this. In other words, what were your calculations for Q and P at an arbitrary time t?
Let's say at time t the position of P is p and the position of Q is q so in terms of p and q and t I can give the new position of P as
r=r0+vt
= (i+7j) + (5i-5j)t
=i+7j + 5it-5jt
=(1+5t)i + (7+5t)j

And new position for Q as
r=r0+vt
=(3i-8j) + (6i+5j)t
=3i-8j + 6it+5jt
=(3+6t)i + (5t-8)j
 
  • #26
LCKurtz
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Is there some reason you don't simplify those? Anyway, now you have them, what next? There was more in post #20. Tell us what you are thinking or why you are stuck.
 
  • #27
lioric
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Is there some reason you don't simplify those? Anyway, now you have them, what next? There was more in post #20. Tell us what you are thinking or why you are stuck.
Well you can see how I have drawn the vector diagram I drew it true to the i j to x y vector coordinates. I just don't know how to go about the question. Some hints would be helpful
 
  • #28
LCKurtz
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You have had some hints. In post #20 I asked you about Q - P. Have you even calculated that yet? Are you going to show us what you get? Have you thought about whether it points Northeast? Or how you would tell? Neither I nor anyone else on this board is going to work it for you.
 
  • #29
You don't need any diagrams. The key is to subtract the vector velocity and position of P from Q. This makes P the new origin and gives you a new location and new vector velocity for Q. You can now use those velocity components to figure out when the two position components will be equal which is when the position of Q is north-east of P.
 
  • #30
lioric
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You don't need any diagrams. The key is to subtract the vector velocity and position of P from Q. This makes P the new origin and gives you a new location and new vector velocity for Q. You can now use those
velocity components to figure out when the two position components will be equal which is when the position of Q is north-east of P.
Thank you all for your help
So the position of
P = (i+7j) with v= (5i-5j)
Q=(3i-8j) with v=(6i+5j)
Already Q is on the east side of P and the rate at with Q moves in x axis is faster than P
So therefore Q will always be on the east side of P
When we look at the y axis it seems that P is moving down at the same rate as Q is moving up
So the moment these two meet in y plan would be the time that Q is exactly on the east of P
So using speed = distance /time
Distance= 7-(-8)= 15
To find mid position 15/2 =7.5
Speed = 5
Time=distance/speed
=7.5/5=1.5hrs
At this time the two ships will be is line in the
With a gap in the x axis
Anything more than 1.5 hours and Q will be in the NE section of P

How's this?
Ok?
 
  • #31
LCKurtz
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How's that? Not good. You haven't even used ##t##.

If you had actually read and responded to my posts you would have had this problem solved a week ago. You haven't simplified your answer in post #25 or responded to post #28. Until you do you will hear nothing more from me.
 
  • #32
lioric
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How's that? Not good. You haven't even used ##t##.

If you had actually read and responded to my posts you would have had this problem solved a week ago. You haven't simplified your answer in post #25 or responded to post #28. Until you do you will hear nothing more from me.
I'm sorry about that
Was away for some time and no internet there
But I'll start working on this now.
Ok so simplifying the new position of P
(1+5t)i + (7+5t)j
P= i+5ti + 7j+5tj
New position of Q
(3+6t)i + (5t-8)j
Q= 3i+6ti + 5tj-8j

Q-P
(3i-8j)-(i+7j)
=(2i-15j)
 
  • #33
LCKurtz
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Make sure you are using the correct numbers as given in post #18.
When you "simplify" a vector it should look like (...)i + (...)j, one i term and one j term.
Your final answer for Q-P at time t must have a "t" in it.
Then we can talk about whether it points NE.
 
  • #34
lioric
280
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Post 18

At noon two boats P and Q have a position vector (i+7j)km and (3i-8j)km respectively to the origin O. i and j are unit vectors in direction to East and north respectively P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
 
  • #35
lioric
280
5
Make sure you are using the correct numbers as given in post #18.
When you "simplify" a vector it should look like (...)i + (...)j, one i term and one j term.
Your final answer for Q-P at time t must have a "t" in it.
Then we can talk about whether it points NE.
I m very confused
In post #25 it seems that the vectors are in the format that you asked (....)i+(...)j and you asked me to simply them. So I expanded them and now your asking me to revert them back again.
 

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