Help with vector exam questions (M1)

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The discussion revolves around solving vector exam questions related to the motion of two helicopters, P and Q. The participants assist in determining the velocity of helicopter P and finding its position vector over time, leading to the expression p = (60t + 20)i + (35 - 120t)j km. For helicopter Q, the focus shifts to calculating its velocity vector based on a given direction and magnitude, resulting in the velocity of Q being (96i - 72j) km/h. The conversation emphasizes using kinematic equations and unit vectors to derive position expressions. Overall, the thread provides a collaborative effort to clarify concepts in vector motion and kinematics.
CathyLou
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Hi.

I have no idea how to even begin to do this question and so I would really appreciate any help.

Two helicopters P and Q are moving in the same horizontal plane. They are modeled as particles moving in straight lines with constant speeds. At noon P is at the point with position vector (20i + 35j) km with respect to a fixed origin O. At time t hours after noon the position vector of P is p km. When t = 0.5 the position vector of P is (50i - 25j) km. Find

(a) the velocity of P in the form (ai + bj) km/h,

I got this as (60i - 120j) km/h.

(b) an expression for p in terms of t.

Thank you.

Cathy
 
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How does distance relate to velocity and time?

(This seems better suited for Intro Physics than Calc & Beyond--I'll move it.)
 
Last edited:
Ok, thanks.

Would I use v = s/t?

Cathy
 
You should start with the basic equations of kinematics and go on from there.
 
CathyLou said:
Ok, thanks.

Would I use v = s/t?

Cathy

Yes, but here s = final position - initial position...

or you can just use sfinal = sinitial + vt
 
CathyLou said:
Would I use v = s/t?
Yes.

You have the initial position vector. You know how each component changes with time, so you can find the position vector at any time t.
 
Thanks everyone for your help.

Sorry, but I still don't see how I can get an expression for p in terms of t.

Cathy
 
CathyLou said:
Thanks everyone for your help.

Sorry, but I still don't see how I can get an expression for p in terms of t.

Cathy

p is just the final position in the equation (the final position of helicopter P after t amount of time passes):

v = (final position - initial position)/t

solve for final position... then you can plug in initial position and velocity since you know those already...
 
Okay - thanks so much for your help!

I got p = (60t + 20)i + (35 - 120t)j km.

Cathy
 
  • #10
CathyLou said:
Okay - thanks so much for your help!

I got p = (60t + 20)i + (35 - 120t)j km.

Cathy

looks good!
 
  • #11
Thank you.

I'm not sure now how to do this next part. Could you please help again?

At noon Q is at O and at time t hours after noon the position vector of Q is q km. The velcoity of Q has magnitude 120 km/h in the direction 4i - 3j. Find

an expression for q in terms of t.


Thanks so much!

Cathy
 
  • #12
Try to get the velocity of Q as a vector... Use the magnitude of the velocity which is given... along with a unit vector in the given direction...

So the first step is to get the unit vector in the given direction.
 
  • #13
learningphysics said:
Try to get the velocity of Q as a vector... Use the magnitude of the velocity which is given... along with a unit vector in the given direction...

So the first step is to get the unit vector in the given direction.

Would I use Pythagoras?

Cathy
 
  • #14
CathyLou said:
Would I use Pythagoras?

Cathy

Yes... have you studied unit vectors in your class? You divide the vector by its magnitude...
 
  • #15
I'm really not sure what to do.

Could you please show me the steps?

Thank you.

Cathy
 
  • #16
CathyLou said:
I'm really not sure what to do.

Could you please show me the steps?

Thank you.

Cathy

First step is to find the unit vector in the direction of 4i-3j... so first find the magnitude of this... it is \sqrt{4^2 + (-3)^2} = 5

So then the unit vector in the direction of 4i - 3j is \frac{1}{5}(4i-3j) = \frac{4}{5}i - \frac{3}{5}j

Multiply this vector by 120km/h... and that's your velocity vector. So the velocity of Q is:

120(\frac{4}{5}i - \frac{3}{5}j) = (96i -72j) km/h

Now you can use this velocity to find q... using the fact that the initial position = 0i + 0j... so now it is just like the previous problem...

In general when you need a vector of a particular magnitude in a particular direction... you multiply the magnitude by a unit vector in that particular direction...

Hope this helps...
 
  • #17
Thank you so much for your help!

Cathy
 

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