# Converting Distance- Time Graph to Velocity- Time

1. Aug 27, 2010

### clamchowder

1. The problem statement, all variables and given/known data
I'm trying to do problem d on page 6.

3. The attempt at a solution

I divided the distance by the time to get each velocity and graphed it. When I took the slope of the line, however I ended up with 4.7. Isn't the answer supposed to be 9.80 if we're calculating the acceleration due to gravity?

2. Aug 27, 2010

### danielakkerma

Hi,
You've got to bear in mind that your object starts from rest, then begins to travel with a constant acceleration, that is, its equation of motion is dictated by:
(1):$$x=\frac{at^2}{2}$$
Therefore:
You can find, a-the acceleration by pasting in what is known(x being D, and t well, is just plain old "t").
Daniel

3. Aug 27, 2010

### clamchowder

I didn't think of it that way, but I'll try it out now.

Thanks!

4. Aug 27, 2010

### clamchowder

I'm still confused on why my slope on my v vs t graph isn't anywhere near 9.80 though.

Distance of Fall (m) .10 .50 1.00 1.70 2.00
Time of Fall (s) .14 .32 . 46 .59 .63
Velocity (D/T) (m/s) .714 1.56 2.17 2.88 3.17

5. Aug 27, 2010

### danielakkerma

Well, look here:
If the equation is given by $$x=\frac{at^2}{2}$$, then clearly,
(1):$$a=\frac{2x}{t^2}$$
(2) That causes the velocity to become $$v=at$$
You're wrong to assume that the slope is simply (D/t). The relationship provided by this equation(1) is parabolic not linear, and you can see that on the first graph they provided you with.
Try computing 'a' with (1). all answers may vary up to ~0.2 and usually center around 10, that's the preferred rounding of 'g' in general.
Daniel

6. Aug 27, 2010

### clamchowder

Oh, I'm sorry.
I never considered the fact it wasn't linear, which was stupid...
At first I was confused where the equations came from, but now I realized it was from
v= v_0 +at
y=.5at^2+v_0t

I finally got the right slope!

Thank you :)