# Stupid Mathematics Issue

1. May 9, 2010

### forest125

1. The problem statement, all variables and given/known data

Okay I've basically solved this problem... but I need some clarification.

The problem is "The acceleration of an object is given as a function of its position in feet by a=2*s^2 ft/s^2. When s=0, it's veloctiy is 1 ft/s. What is the velocity of the object when s=2 ft?

2. Relevant equations

So since a=2*s^2, we can say dv/dt=2*s^2, and dv/dt=dv/ds*ds/dt=dv/ds*v

3. The attempt at a solution

Well, since acceleration is dv/ds*v, Int(v*dv)=Int(2*s^2*ds)
So after integration, v^2/2=2/3*s^3.

So upon solving for v, v^2=4/3*s^3, and therefore v=sqrt(4/3*s^3). With the initial condition when s=0, v=1, v=sqrt(4/3*s^3)+1 is what I get.

This is wrong though. The correct answer to the problem is 3.42 ft/s, which I can get when v=sqrt(4/3*s^3+1), in other words, just including the 1 in the square root. I know I'm likely doing something really stupid...

All I can figure is that like v^2/2=2/3*s^3+c... where c=1, only trouble is I don't think that's a true statement.

So anyway... sorry for excrutiating detail, I just CANNOT figure out where I'm going wrong here.

Thanks in advance. :)

2. May 9, 2010

### forest125

Ok. I think I'm over thinking this. v^2/2=2/3*s^3+c, so v^2=4/3*s^3+2c, and then v(s)=sqrt(4/3*s^3+2c), and since v(0)=1, sqrt(2c)=1, therefore 2c must=1. Is that the right way to think about this?

Again sorry, I know these are really easy concepts we're dealing with here. Just wanna make sure I understand the algorithm to solve these simple diff eq's in dynamics situations.

3. May 9, 2010

### Dick

v^2/2=2/3*s^3+c is figuring it correctly. You add the constant when you integrate. You don't add the constant after you've solved for v.