Constant Acceleration in a Calculus Word Problem

[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

 

[e(ho0n3]

Isn't ds/dt just v?

 

[Hootenanny]

Isn't ds/dt just v?

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

 

[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.

 

[Hootenanny]

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.

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

~H

 

[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?

 

[Hootenanny]

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

 

[preet]

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

-Preet