# Acceleration in function of time and distance

by sinnet3000
Tags: acceleration, distance, function, time
 P: 2 Here is something that has got me confused a lot of times. Suppose I have a distance of 11m and a time of 5 seconds and I want to know acceleration. I would say that $a = v / t$ and $v = d / t$ so I could plug the second equation to the first equation having: $a = d / t^2$ Therefore I have $a = 11 / 5^2m/s^2$ but that is the wrong answer. We also have $x=x_0+v_0 t+1/2 at^2$ so: $a = 2d / t^2$ which in that case is $a = 2(11) / 5^2$ and this is right. But why is the reason that distance should be the double of it?? Can someone explain me please?? Thank you