VCAA 2007 Exam 2, Question 4. Could someone please post a worked solution?

[ahat]

I'm having trouble understanding parts C and D
It would be much appreciated, thanks very much
(it would be great if you could explain why the method was used rather than just posting it. for example. for d, I know we find the dot product of position and velocity vector, I'm not sure why it's those two specifically)

Kind regards in advance!
(PLEASE READ THE SECOND ATTACHMENT - full question)

[bonappler]

Okay so if you want the angle then you must use the velocity, find the modulus of the vector and this will equal the hypotenuse, 4 will be opposite as it is the height of the plane, then use sin to find the angle. For part d you can find the magnitude of the position vector, differentiate this, equate to 0 and and then solve for t.
I suppose you can use cos as well but this would be slightly longer as you have to resolve i and j to get the adjacent side.
I hope this helped.

[kaybee94]

I hope the working out is clear! 

[nspire]

I hope the working out is clear! 

Jack's Tuition Springvale....

[ahat]

I hope the working out is clear! 

Thanks so much man! It was really helpful - especially part d! Ahh, that was such a great method (makes super sense).

For part e though, I preferred to use this method: |r(60) - r(0)|
Originally though, like many other students apparently, I did |r(60)| - |r(0)|. Can someone pleaseee explain why this is wrong, it's essential to improving my understanding.

And I apologise for asking this question again, but why is it that we use the velocity vector for part c? Why not r(t)? Or (this may sound silly), how do we know that the velocity vector is going to model the plane landing anyway?

[kaybee94]

in the answer it said that "direction corresponds to velocity vector". SInce it is moving then u use velocity vector rather than ur r(t) which is just position from ur original starting position. To answer ur other q, the reason is if ur trying to find the distance from t=0 to t=60 u take the displacement between the two and find the magnitude. Taking the magnitude of each one separately and then finding the difference is incorrect.  |r(60)| - |r(0)| is saying ur finding the difference between each point with respect to the original position rather than the direct distance from t=0 to t=60 which is the other formula. The easier method is obviously to find magnitude of velocity x time.

Jack's Tuition Springvale....

Yes it is he explained it a lot better than vcaa solutions imo.

[ahat]

Thankyou kaybee

[duhherro]

Hi sorry for this bump, but going over the exam, is the |r(t)| showing the direction of the plane flight , where x-axis would be considered the control tower? So therefore, a minimum TP would be closest to the x-axis (control tower)

[Sentar]

Hi sorry for this bump, but going over the exam, is the |r(t)| showing the direction of the plane flight , where x-axis would be considered the control tower? So therefore, a minimum TP would be closest to the x-axis (control tower)

|r(t)| is the distance from the control tower, which is located at the origin (0,0,0), as it is the magnitude it doesn't show direction. The minimum TP does give the smallest distance between the plane and the tower as the distance decreases and then begins increasing again once the plane passes the tower.

[duhherro]

|r(t)| is the distance from the control tower, which is located at the origin (0,0,0), as it is the magnitude it doesn't show direction. The minimum TP does give the smallest distance between the plane and the tower as the distance decreases and then begins increasing again once the plane passes the tower.

Thanks a lot!