Motion Question from HSC 98 Q6bii) (1 Viewer)

[clintmyster]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/98MATH34.PDF

Can anyone provide me a solution to this question? I got the solution from success one but I don't think it is adequate enough to gain two marks.

Any help would be much appreciated.

 

[xFusion]

i would just put a root around the x values on the right and say 'no, because the velocity is always positive' lol i dunno if thats right

 

[clintmyster]

it actually does go back through the origin.

 

[xFusion]

it actually does go back through the origin.

lol thats a sign that i should review this topic before my trials on friday.

 

[Trebla]

a = d(v²/2)/dx = 3(1 - x²)
v² = 16 + 6x - 2x³

Let y = v²
.: dy/dx = 6(1 - x²)
.: y = 16 + 6x - 2x³
At x = 1, y = 20 (max)
At x = -1, y = 12 (min)

The particle initially moves in the positive direction (as indicated by its positive velocity). When x > 1, the acceleration is always negative thus it slows down after x = 1 and will eventually come to rest for a moment (at the position say x₁ which satisfies 16 + 6x₁ - 2x₁³ = 0) before going back in the other direction (velocity changes from positive to negative after passing the "zero") and therefore return to the origin.
We can see that it doesn't turn around again before it hits the origin because v cannot be zero again since the y-values of turning points of y = 16 + 6x - 2x³ are the same sign (a sketch of y = 16 + 6x - 2x³ should confirm this).

Hopefully that makes sense...

 

[clintmyster]

thanks trebla, that does make sense. Do you need all of that for 2marks tho?

 

[blaze.bluetane]

whats the success one solution?