What is the Problem with Two Ships P and Q in Constant Motion?

[CathyLou]

Hi.

Could someone please help me with the following question? I've no idea how to begin

Two ships P and Q are moving along straight lines with constant velocities. Initially P is at a point O and the position vector of Q relative to O is (6i + 12j) km, where i and j are unit vectors directed due east and due north respectively. The ship P is moving with velocity 10j km/h and Q is moving with velocity (-8i + 6j) km/h. At time t hours the position vectors of P and Q relative to O are p km and q km respectively.

(a) Find p and q in terms of t.

Thank you.

Cathy

 

[learningphysics]

Does it say p and q are the magnitudes of the position vectors?

 

[berkeman]

You have initial position vectors and velocities. Express the relative motion in rectangular coordinates.

 

[CathyLou]

Does it say p and q are the magnitudes of the position vectors?

No. It just says position vectors.

Cathy

 

[CathyLou]

You have initial position vectors and velocities. Express the relative motion in rectangular coordinates.

What is the initial position vector for P?

Cathy

 

[learningphysics]

No. It just says position vectors.

Cathy

Oh... I think I was just misunderstanding... So p is the final position of ship P and q is the final position of ship Q (all relative to 0)...

you know that \vec{s} = \vec{s_{0}} + \vec{v}t

where \vec{s} is the final position. Use this formula to find the final position of P and Q... treat P and Q as separate problems... one really has nothing to do with the other...

So what is \vec{s_{0}} and \vec{v} for P?

What is \vec{s_{0}} and \vec{v} for Q?

 

[learningphysics]

What is the initial position vector for P?

Cathy

It is just 0. Or 0i + 0j.

 

[CathyLou]

It is just 0. Or 0i + 0j.

Oh, thanks very much. I understand now!

Cathy

 

[learningphysics]

Oh, thanks very much. I understand now!

Cathy

No prob.

 

[CathyLou]

Oh yikes - I'm now stuck on these two parts!

Could someone please give me some hints? I would really appreciate any help!

Calculate the distance of Q from P when t = 3.

Is t measured in seconds, minutes or hours?

Calculate the value of t when Q is due north of P.

Thank you.

Cathy

 

[learningphysics]

t is in hours according to the question description... did you calculate the formulas for p and q?

Do you have any ideas of how to approach the problem? Hint: Use your position formulas p and q

 

[CathyLou]

t is in hours according to the question description... did you calculate the formulas for p and q?

Do you have any ideas of how to approach the problem?

Oh yeah, of course it is!

Yup, I got p = 10jt and q = (6 - 8t)i + (6t + 12)j.

Would it be possible to work out the displacement and then use v = d/t?

Cathy

 

[learningphysics]

The formulas look good. Yeah, you need the displacement from P to Q. But you don't need velocity. You need distance... so get the distance from your displacement.

 

[CathyLou]

Thanks so much.

I got that the distance is 18 km.

Am not sure how to do the next part, though.

Could you please help?

Thank you.

Cathy

 

[learningphysics]

Thanks so much.

I got that the distance is 18 km.

Am not sure how to do the next part, though.

Could you please help?

Thank you.

Cathy

The distance looks good to me. :)

For the next part. Use your displacement formula from P to Q which is in terms of t...

Hint: you want to find the time when the i component of the displacement 0. when that happens, either Q is north or south of P. you should check the number you get to make sure Q is north and not south of P.

 

[CathyLou]

Thanks for your help.

I got that t = 0.75 hours.

I'm not sure how to check the number to check that it is north and not south of P. Could you please explain this?

Cathy

 

[learningphysics]

Cool. That's what I get.

In your displacement formula from P to Q, when you substitute in t = 0.75, you should get a positive coefficient for j... that means that Q is north of P... if it was negative Q would be south of P.

I get the displacement from P to Q at t=0.75 as q - p = 9j, so since 9>0 Q is north of P...

 

[CathyLou]

Okay. I get it now!

Thanks again for all your help.

I really appreciate it! :-)

Cathy

 

[learningphysics]

Okay. I get it now!

Thanks again for all your help.

I really appreciate it! :-)

Cathy

You're welcome.