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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: elle.123 on October 27, 2009, 10:51:38 pm

Title: Velocity question
Post by: elle.123 on October 27, 2009, 10:51:38 pm: How do you tell whether v is meant to be positive or negative with questions like these?

Question: A particle initially at rest at an origin, moves along a straight line with a velocity v. The particle has an acceleration of (3+2x)^-2, where x is its position from the origin.

I have got to v= ±√((-1)/(3+2x)), and i don't know how to figure out if v is meant to be positive or negative.

TIA.
Title: Re: Velocity question
Post by: Mao on October 28, 2009, 12:39:09 am: Initial acceleration is $a(0) = (3)^{-2} = \frac{1}{9}$ , this is in the positive x direction. Hence the object moves from origin in the positive x direction, :. v is positive.

However, I do believe your expression for v is incorrect:

$a = \frac{d}{dx} \left( \frac{1}{2}v^2\right) = (3+2x)^{-2}$

$\implies \frac{1}{2}v^2 = -\frac{1}{2(3+2x)} + C$ , given $t=0,\; v=0,\; \implies C = \frac{1}{6}$

$\implies v = \sqrt{\frac{1}{3}-\frac{1}{3+2x}}$ taking the positive branch.
Title: Re: Velocity question
Post by: plato on October 28, 2009, 12:43:51 am: Quote from: elle.123 on October 27, 2009, 10:51:38 pm
How do you tell whether v is meant to be positive or negative with questions like these?

Question: A particle initially at rest at an origin, moves along a straight line with a velocity v. The particle has an acceleration of (3+2x)^-2, where x is its position from the origin.

I have got to v= ±√((-1)/(3+2x)), and i don't know how to figure out if v is meant to be positive or negative.

TIA.

Your integral is not quite complete as you have forgotten the constant of integration. Without it, you would not get v=0 at x=o as stated in your question.

You should get $v^2=-\frac{1}{3+2x}+c}$
Then, substituting v=0 at x=0, you get c = $\frac{1}{3}$
and so $v^2=\frac{1}{3}-\frac{1}{3+2x}$

Then $v=\pm\sqrt{\frac{1}{3}-\frac{1}{3+2x}}$

From here, consider that the acceleration function is always positive (due to the squaring effect of the power) and tending toward a value of zero.
Therfore the velocity must be in the direction of the acceleration and tending toward a limiting value of $v=\sqrt{\frac{1}{3}$ as x gets larger.