What Is the Optimal Distance D for Minimum Time on a Frictionless Track?

[Shadowness]

Hello all,

I never really had a great deal of trouble with physics until just recently and this place looks like a great place to learn. I am having problems with the following question:

A frictionless track is to be built as shown, with L=3.20 m and H=2.90 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.
Hint: A minimum or maximum value can be found using derivatives.
http://www.shadowsillusion.com/images/ramp2.gif

How would I go about in setting this problem up? I am just clueless one where to start. I believe that I have to set up some type of formula, take the derivative of it and solve for D?

Thank you for your time and help.

 

[OlderDan]

Some hints:

No friction means energy is conserved.
Acceleration for frictionless inclines is constant
Average velocity is easy to calculate when acceleration is constant
Average velocity is distance divided by time.
Total path length is a fairly easy trig problem

Tell us what you think needs to be done

 

[thepatientmental]

Hi there,

Newton's Laws of Motion are fundamental principles in physics that explain how objects move and interact with each other. They have many practical applications in our daily lives, including in the design and construction of structures and machines, as well as in understanding the motion of celestial bodies.

In the case of the problem you mentioned, Newton's Laws can be used to determine the minimum time it takes for a cart to slide down a frictionless track. The first law, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will continue in motion with a constant velocity unless acted upon by an external force. This means that once the cart is set in motion, it will continue to move down the track without any external forces acting on it.

The second law, also known as the law of acceleration, states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that the larger the force pushing the cart down the track, the faster it will accelerate. In this case, the force pushing the cart down the track is its weight, which is equal to its mass multiplied by the acceleration due to gravity.

The third law, also known as the law of action and reaction, states that for every action, there is an equal and opposite reaction. This means that the force of the track pushing up on the cart is equal and opposite to the force of the cart pushing down on the track.

To solve the problem, we can use the equation F=ma, where F is the force, m is the mass, and a is the acceleration. We can also use the equation for the acceleration due to gravity, a=g, where g is approximately 9.8 m/s^2. We can set up the problem as follows:

F=ma
mg=ma
g=a

Using the third law, we know that the force pushing the cart down the track is equal to its weight, which is mg. We also know that the acceleration of the cart is equal to the acceleration due to gravity, which is g. Therefore, we can plug in these values into the equation for acceleration and solve for g.

g=9.8 m/s^2

Next, we can use the equation for displacement, d=vt+1/2at^2, where d is the displacement, v is the initial velocity, t is the time, and a