Calculus: Given parabola and vx in terms of x and c, prove constant velocity

[derelictee]

Here an AP Physics problem that's really bugging me.

Homework Statement

A particle moves along the parabola with equation y = .5x^2

part a) I believe I did this correct.

part b) Suppose that the particle moves with a velocity whose x-component is given by vx = c / (1 + x^2)^.5 Show that the particle's speed is constant.

Below I have images of the question and my attempted work. I think maybe for the first half of my work I was in the right direction; I got y in terms of t, and I was going to find the derivative to show that there is no acceleration, but I couldn't get the equation to equal y, and I ultimately became confused and went off track.

The Attempt at a Solution

http://img363.imageshack.us/img363/7313/scanqa9.th.jpg http://g.imageshack.us/thpix.php
http://img218.imageshack.us/img218/2346/scan0001fu2.th.jpg http://g.imageshack.us/thpix.php
I know; my work is a mess.

 

[LowlyPion]

Here an AP Physics problem that's really bugging me.

Homework Statement

A particle moves along the parabola with equation y = .5x^2

part a) I believe I did this correct.

part b) Suppose that the particle moves with a velocity whose x-component is given by vx = c / (1 + x^2)^.5 Show that the particle's speed is constant.

Below I have images of the question and my attempted work. I think maybe for the first half of my work I was in the right direction; I got y in terms of t, and I was going to find the derivative to show that there is no acceleration, but I couldn't get the equation to equal y, and I ultimately became confused and went off track.

I know; my work is a mess.

Have you considered that if Vx = c/(1+x²)^1/2 that x would be given by the integral?

In this regard isn't the x position as a function of time given by the integral of Vx(t)

$$\int \frac{c*dx}{(x^2 + 1)^{1/2}} = c*arcsinh(x) + C = c*arcsinh(x) + c$$

C is c because V=c at x=0

 

[baba89]

I would respond by saying that your approach in part a) was correct, however, your method in part b) was not entirely correct. In order to prove constant velocity, you need to show that the particle's acceleration is zero. In this case, you can use the equation for velocity, vx = c / (1 + x^2)^.5, to find the acceleration, ax, by taking the derivative with respect to time, t.

ax = d(vx)/dt = -cx / (1 + x^2)^1.5

Since the acceleration is dependent on x, we need to express x in terms of t using the equation for the parabola, y = .5x^2. This can be done by solving for x in terms of y, and then substituting that expression into the equation for acceleration.

x = √(2y)

ax = -c√(2y) / (1 + 2y)^1.5

Now, we can see that the acceleration is dependent on y, not t. This means that the acceleration is constant and does not change over time. Therefore, the particle's acceleration is zero and its speed is constant. In other words, the particle is moving with a constant velocity.

I hope this helps to clarify the solution for part b) of the problem. Keep up the good work in your AP Physics class!