A frictionless track is to be built as shown, with L= 4.6 m and H= 3.9 m. In order to get the cart to slide from the top to the end of the track in the minimum time, how long should the distance D be? Assume that the speed of the cart on the horizontal surface is the same as at the bottom of the ramp.
http://newcapa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/11_Force_Motion_Adv/graphics/prob05_ramp2.gif

Um... I don't see any track 'as shown' :(

Can you give a description of it?

Here is it :
http://img836.imageshack.us/img836/3479/prob05ramp2.gif

Can u see the pivture now?

Yes, but I'm not sure how to solve this in a simple way. I got something but it seems too complicated. I'm still thinking about it.

And one thing that bugs me is this line:
"Assume that the speed of the cart on the horizontal surface is the same as at the bottom of the ramp. "

What if D = 0?
The speed at the bottom of the ramp will be maximum, but then, it will have to horizontal component so that the cart can move on the horizontal track...

Bah, anyway, this is what I did.

The acceleration of the cart is g\sin\theta

Along the horizontal, this is: g \sin\theta \cos\theta = 0.5gsin2\theta

The horizontal speed at the bottom of the ramp is v^2 = 2(0.5gsin2\theta)(D) = gD\sin2\theta

The time it takes to cover D is t_1 = \frac{\sqrt{gD\sin2\theta}}{0.5gsin2\theta} = \frac{2\sqrt{gD\sin2\theta}}{gsin2\theta}

To cover the rest, it takes:

t_2 = \frac{4.6 - D}{\sqrt{gD\sin2\theta}}

Total time:

T = \frac{2\sqrt{gD\sin2\theta}}{gsin2\theta} + \frac{4.6 - D}{\sqrt{gD\sin2\theta}}

This simplifies to:

T = \frac{D + 4.6}{\sqrt{gD\sin2\theta}}

From the picture, we know that \tan\theta = \frac{3.9}{D}

Therefore, \sin\theta = \frac{3.9}{\sqrt{3.9^2 + D^2}} and \cos\theta = \frac{D}{\sqrt{3.9^2 + D^2}}

Hence, 2\sin\theta \cos\theta = 2\(\frac{3.9}{\sqrt{3.9^2 + D^2}}\)\(\frac{D}{\sqrt{3.9^2 + D^2}}\) = \frac{7.8D}{3.9^2 + D^2}

Using this in the previous equation;

T = \frac{D + 4.6}{\sqrt{gD\(\frac{7.8D}{3.8^2 + D^2}\)}} = \frac{(D + 4.6)(\sqrt{3.9^2 + D^2})}{D\sqrt{7.8g}} = \frac{(D + 4.6)(\sqrt{15.21 + D^2})}{8.74D}

Now, to find the minimum time, we have to differentiate this using first the quotient rule, then the product rule.

T' = \frac{(8.74D)\[(D+4.6)(0.5(15.21 + D^2)^{-0.5}(2D)) + (15.21 + D^2)^{0.5}(1)\] - (D + 4.6)(15.21 + D^2)^{0.5}(8.74)}{76.44D^2}

And solve for 0.

0 = 8.74D\(\frac{D^2+4.6D}{\sqrt{15.21 + D^2}} + \sqrt{15.21 + D^2}\) - (8.74D + 40.2)\sqrt{15.21 + D^2}

(8.74D + 40.2)\sqrt{15.21 + D^2}= 8.74D\(\frac{D^2+4.6D}{\sqrt{15.21 + D^2}} + \sqrt{15.21 + D^2}\)

(8.74D + 40.2)(15.21 + D^2) = 8.74D\(D^2+4.6D + 15.21 + D^2\)

133D + 8.74D^3 + 611 + 40.2D^2 = 17.5D^3 + 40.2D^2 + 133D

0 = 8.74D^3 -611

D^3 = \frac{611}{8.74} = 69.966

Note: You get 69.966 only if you saved all the decimals

D = 4.12\ m

Well, that's what I get... I know it's very long though...

And I checked my derivative just in case I messed up something, but it is good:

http://www.wolframalpha.com/input/?i=min{y+%3D+\frac{(x+%2B+4.6)(\sqrt{15.21+%2B+x^2 })}{8.74x}}

this site gives the local minimum at 4.12 m, just like I got.