Question

What is d=1/2gt^2?

Answer 1

if an object is dropped from a distance d with zero initial velocity and it takes time t to reach the ground then....
\[d=1/2g t ^{2}\]
where g is the gravitational acceleration...

Answer 2

um sorry, but can you please explain why?

Answer 3

\[d =\frac{ 1 }{ 2 }g t^{2} \]

Answer 4

velocity is the rate of change in distance per time.

acceleration is the rate of change in velocity per time

using calculus, the derivative of distance tells us the instantaneous velocity; and the derivative of velocity tells us the instantaneous acceleration:

spose an object is simply not moving, its distance is simply constant: the derivative of a constant is 0

spose an object IS moving such that its distance with respect to time is linear (a constant velocity): the derivative of a line function produces a constant: \[y = kt\\\frac{dy}{dt}=k\]and since the derivative of a constant is zero, the object is simply not accelerating.

when we work this out with a constant acceleration, we end up with a starting point of distance as:\[\frac12a~t^2\]
if we plot the distance with respect to time of an object with constant acceleration, it forms a parabolic curve, a quadratic.