How to find slowing acceleration from position versus time graph and the two equations
x-xnaught=vnaught + (1/2)t^2
x=At^2+Bt+C

Question

How to find slowing acceleration from position versus time graph and the two equations
x-xnaught=vnaught + (1/2)t^2
x=At^2+Bt+C

anonymous · Answer

my teacher said I had to use the velocity vs. time graph, where velocity slope was decreasing, and take best fit curve. from that A=1/2a so acceleration is 2A. why is that?? confusing!

anonymous · Answer

actually from first equation it is 1/2at^2

ash2326 · Answer

oh I have an idea now
we have the curve$$x=At^2+Bt+C$$
this is equation of a parabola
Have you learnt calculus?

ash2326 · Answer

or you can find the slope of the velocity time curve, which will be equal to acceleration.
Since rate of change of velocity is acceleration.

anonymous · Answer

here is ws. I think that may work. but what is she saying about taking 2A is equal to acceleration fro quadratic eq. above?

anonymous · Answer

ok thank you

anonymous · Answer

x-xnaught=vnaught + (1/2)t^2 how was this derived though? do you know?