# Thread: Im completely stumped by this one.

1. ## Im completely stumped by this one.

Find an equation for the distance traveled by an object moving along a straight line if the velocity at time t is given by: v=1/the square root of (9-t^2). The object was 10m from the point of reference at t=0s.

I think you need to use integrals with inverse trigonometry, but im not sure. Please help!!!

2. Note that $s = \int v dt$ where s represents distance. So:

$s = \int \frac{1}{\sqrt{9 - t^{2}}}dt$

Do a substitution of:
$t = 3\sin \theta$
$dt = 3\cos \theta d\theta$

So you get:
$s = \int \frac{3\cos \theta d\theta}{\sqrt{9 - 9\sin^{2} \theta}} = \int \frac{3\cos \theta d\theta}{3 \sqrt{1 - \sin^{2} \theta}}$

Can you see where you can go with the $1 - \sin^{2} \theta$?

Note that there will be the constant after integrating. They give you the point (0, 10) to allow you to solve for it.

3. Originally Posted by o_O
Note that $s = \int v dt$ where s represents distance. So:

$s = \int \frac{1}{\sqrt{9 - t^{2}}}dt$

Do a substitution of:
$t = 3\sin \theta$
$dt = 3\cos \theta d\theta$

So you get:
$s = \int \frac{3\cos \theta d\theta}{\sqrt{9 - 9\sin^{2} \theta}} = \int \frac{3\cos \theta d\theta}{3 \sqrt{1 - \sin^{2} \theta}}$

Can you see where you can go with the $1 - \sin^{2} \theta$?

Note that there will be the constant after integrating. They give you the point (0, 10) to allow you to solve for it.
Uhh....trig substitution...cant this be done using simple integration techiniques with arcin? $\int\frac{1}{\sqrt{9-t^2}}dt=arcsin\bigg(\frac{t}{3}\bigg)+C$...?

4. Oh whoops. Thought he said he wanted a trig sub.

5. Originally Posted by o_O
Oh whoops. Thought he said he wanted a trig sub.
Haha...at least he can now see where the formula comes from