# Calculus word problem

1. Apr 18, 2006

### preet

"An object moves so that its velocity, v is related to its position. s according to v = (b^2 + 2gs) ^1/2 where b and g are constants. Show that the acceleration of the object is constant."

I typed out the question exactly as it is. I'm confused because I don't really get what to do. To show that acceleration is constant, I need to get rid of that "s" variable in the question. Acceleration is = to d(velocity)/dt... but from the given function, you can only get d(v) / ds.

So dv/dt = dv/ds * ds/dt

But how do I find d(s) / dt? Don't I need a function that has position in terms of time?

TiA
Preet

2. Apr 18, 2006

### e(ho0n3

Isn't ds/dt just v?

3. Apr 18, 2006

### Hootenanny

Staff Emeritus
Yes but v isn't a function of t in this case, it is a function of s

4. Apr 18, 2006

### e(ho0n3

Does that matter? I mean, a = dv/ds * ds/dt = v dv/ds. If it can be shown that this is constant, then the problem is solved. There is no need to get rid of s.

5. Apr 18, 2006

### Hootenanny

Staff Emeritus
Could you please show me how;

$$\frac{dv}{ds} \cdot \frac{ds}{dt} = v\frac{dv}{ds}$$

Perhaps I'm missing something? It is late after all and I've run out of coffee :grumpy:

~H

6. Apr 18, 2006

### e(ho0n3

dv/ds * ds/dt = ds/dt * dv/ds by commutativity and substituting ds/dt for v gives the result I gave.

Is there a flaw in my reasoning here?

7. Apr 18, 2006

### Hootenanny

Staff Emeritus
Ahhhh It was so simple I missed it . My frantic scribblings on paper seem so stupid now.

$$\frac{ds}{dt} = v$$

You reasoning is perfect e(ho0n3.

To clarify for the OP;

You were right by using the chain rule to obtain;

$$\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}$$

What you (and I) didn't spot is that;

$$\frac{ds}{dt} = v \Rightarrow a = \frac{dv}{dt} = v\frac{dv}{ds}$$

As e(ho0n3 correctly stated.

~H

Last edited: Apr 18, 2006
8. Apr 18, 2006

### preet

Yeah, I feel bad not catching that... it was so obvious =/. Thanks a lot for the help!

-Preet