Converting Distance- Time Graph to Velocity- Time

AI Thread Summary
The discussion centers on solving a physics problem related to converting a distance-time graph to a velocity-time graph. The initial confusion arose from calculating the slope of the velocity graph, which yielded an incorrect value of 4.7 instead of the expected 9.80 m/s² for gravitational acceleration. Participants clarified that the motion is governed by a parabolic relationship, not linear, and provided the relevant equations to derive acceleration. After understanding the correct approach, the original poster successfully recalculated the slope. The conversation emphasizes the importance of recognizing the nature of the motion when interpreting graphs in physics.
clamchowder
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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
 
I didn't think of it that way, but I'll try it out now.

Thanks!
 
danielakkerma said:
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

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
 
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
 
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 :)
 
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