Question

Hey all,

I've been puzzling over this one for awhile now.

When we graph velocity (distance/time) versus time, and we take the derivative of this function, we get an acceleration function. If we graph position (just distance) versus time squared, and we take the derivative - do we still get an acceleration function...?

Answer 1

This link says yes: http://wiki.answers.com/Q/What_does_the_slope_of_a_distance_vs_time_squared_graph_represent *but we only get half the acceleration value* - can anyone please explain why?

Answer 2

you integrate acceleration twice to get displacement, if you're dealing with constant acceleration (which you are) 'a' is a constant:
\[d = \int\limits_{ }^{ }\int\limits_{ }^{ }a dt = \frac{ a }{2 }t ^{2}\]

Answer 3

Thank you! :) Hmmm... sorry, but I'm not quite seeing yet how this step leads to the 1/2 difference in acceleration values for the displacement vs time^2 and velocity vs time graphs. Would you mind explaining a bit more, please?

Answer 4

Derivative of position (displacement) is velocity.
Derivative of velocity is acceleration.
Integral of acceleration is velocity.
Integral of velocity is displacement.

Answer 5

Oh, sorry, just noticed the 'distance vs. time-squared' bit...

Answer 6

Distance vs. time squared should be linear for constant acceleration (if I'm considering this correctly.)

Answer 7

Yes - our graph of distance vs time squared was linear.

Answer 8

However, the slope of this graph was, say, k.

The slope of a different graph (using the same data, but plotted to yield velocity vs time instead) resulted in 2k.

Just curious to why that is

Answer 9

plot d vs t^2/2 to get a line with the slope of 'a'