# time, velocity, position of a particle

• Mar 7th 2009, 08:21 AM
LexiRae
time, velocity, position of a particle
at t time a particle is moving along the x axis at position x. the relationship between x and t is given by tx=x^2 +8. at x = 2 the velocity of the particle is:
• Mar 7th 2009, 08:38 AM
Mentia
Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.

Then:

$\displaystyle t=x^{2}+8$

then

$\displaystyle x=\pm \sqrt[ ]{ t-8}$

then

$\displaystyle velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}$

Then if x = 2, velocity = 1/4

Seem reasonable?
• Mar 7th 2009, 09:09 AM
HallsofIvy
Quote:

Originally Posted by Mentia
Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.

Then:

$\displaystyle t=x^{2}+8$

then

$\displaystyle x=\pm \sqrt[ ]{ t-8}$

then

$\displaystyle velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}$

Then if x = 2, velocity = 1/4

Seem reasonable?

I think simpler: from $\displaystyle t= x^2+ 8$, $\displaystyle 1= 2x dx/dt$. At x= 2, $\displaystyle 1= 4 dx/dt$ so $\displaystyle v= dx/dt= 1/4$.