# Does this work?

1. May 1, 2004

### JasonRox

Here is the info.

s = 150m
a = ?
t = ?
vi = 0
vf = 27

Here is what I did:

$$27 = at$$
$$27/t = a$$

Put a into another unsolved equation.

$$150 = 1/2(27/t)t^2$$
$$11.1 = t$$

Put the known variable into the first equation.

$$27 = a(11.1)$$
$$2.43 = a$$

All I did was take variables, and insert it into it's derivative, or antiderivative. I'm getting different answers using other methods. Can someone explain if this is incorrect?

Last edited: May 2, 2004
2. May 2, 2004

### Integral

Staff Emeritus

I get t=11.1s

3. May 2, 2004

### JasonRox

Yeah, you're right.

Human error.

So, even though you are matching up numbers of different graphs, it still is valid?

4. May 2, 2004

### Integral

Staff Emeritus

Actually you are solving the Differential Equation:

$$\frac {d^2s} {dt^2} =a$$
with initial conditions

$$\frac {ds(T)} {dt}= 27 \frac m s$$

$$\frac {ds(0)} {dt}= 0 \frac m s$$

$$s(T) =150m$$

$$s(0)=0$$

T is the time you need to find.

This can be solved by integration, with each integration constant evaluated using the initial conditions. So your "velocity graph" is the derivative of the "position graph" and are thus simply showing different aspects of the same physical situation.

5. May 2, 2004

### HallsofIvy

The difficulty in telling you if this method works for solving the problem is that you never actually told us what the problem is! You said you did this and that but what is it you are trying to solve?

6. May 2, 2004

### JasonRox

I gave you the info. I had, and showed my steps.

I wasn't really concerned on finding the answer. I was just curious if using certain variables into other equations to find the answer. I know you can do this, but I wasn't quite sure if you can do it with its own derivative.

I guess you can. Thanks for the advice.