Login

Welcome, Guest. Please login or register.

April 27, 2026, 10:54:22 am

Author Topic: VCAA 2011 Exam 2 Question 4  (Read 1663 times)  Share 

0 Members and 1 Guest are viewing this topic.

Patches

  • Victorian
  • Trendsetter
  • **
  • Posts: 131
  • Respect: +23
VCAA 2011 Exam 2 Question 4
« on: November 05, 2012, 08:42:31 pm »
0
Does anyone have a worked solution for Question 4 c. on last year's VCAA Exam 2? The answer in the examiner's report isn't very detailed.

I understand how to get the first half of the final solution for L, but I cannot for the life of me work out the second half, which involves the section where Tasmania Jones moves along the parabola. I've attached a printscreen of the question.

Thanks!

polar

  • Guest
Re: VCAA 2011 Exam 2 Question 4
« Reply #1 on: November 05, 2012, 08:55:53 pm »
+2
« Last Edit: November 05, 2012, 08:58:07 pm by polar »

Patches

  • Victorian
  • Trendsetter
  • **
  • Posts: 131
  • Respect: +23
Re: VCAA 2011 Exam 2 Question 4
« Reply #2 on: November 05, 2012, 09:00:02 pm »
0
Thanks for the quick reply!

That certainly helps, but I'm still struggling from the 6th line of your solution onwards. Are we looking specifically for the distance along the parabolic curve?

Where did you derive the 3/4 - y   from?

polar

  • Guest
Re: VCAA 2011 Exam 2 Question 4
« Reply #3 on: November 05, 2012, 09:02:32 pm »
0
it was the difference between the y-coordinates of the plant and where he reaches - the y-coordinate of the plant is , thus it can be written as

Patches

  • Victorian
  • Trendsetter
  • **
  • Posts: 131
  • Respect: +23
Re: VCAA 2011 Exam 2 Question 4
« Reply #4 on: November 05, 2012, 09:06:08 pm »
0
Ah! I was going in circles trying to find the exact point on the river he would reach and using that instead of just (x,y).

Thank you :)

Lasercookie

  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 3167
  • Respect: +326
Re: VCAA 2011 Exam 2 Question 4
« Reply #5 on: November 05, 2012, 09:10:38 pm »
+1
edit: got beaten by polar (:o so fast at LaTeX polar!!) but typed this up and posting anyway in case anyone likes explanations with words.

We split his adventure up into two. First he runs directly to the point (x,y) on the river from his camp and after that he starts swimming.

So the distance from the origin, where Tasmania has set up camp, to the point (x,y) on the river. The river itself is given by y = x^2 - 1

<-- that's in kilometres (the start of Q4 states that)

How long would it take him to travel that distance? He runs at 2 km per hour.

(so that means T is in hours)

So what about when he actually gets to that point on the river? He starts to swim. We're told that this time is proportional to difference between the y-coordinate of the plant and where he enters the river.

The y-coord of the plant is 3/4, he enters the river at x^2 - 1. It was only proportional to, so we multiply the difference of the two by some constant k.


The total time is then given by


and then expand etc. to get it into the form the question wanted

edit: fixed typo with units
« Last Edit: November 05, 2012, 09:15:07 pm by laseredd »