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susus · Dec 2, 2010, 05:03 PM

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.

Unknown008 · Dec 3, 2010, 12:17 AM

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

Can you give a description of it?

susus · Dec 3, 2010, 04:22 AM

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

susus · Dec 3, 2010, 08:53 AM

Can u see the pivture now?

Unknown008 · Dec 3, 2010, 10:12 AM

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...

Unknown008 · Dec 3, 2010, 11:17 AM

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.