# Integral of Acceleration with respect to time

## Homework Statement

Acceleration is defined as the second derivative of position with respect to time: a = d2x/dt2. Integrate this equation with respect to time to show that position can be expressed as x(t) = 0.5at2+v0t+x0, where v0 and x0 are the initial position and velocity (i.e., the position and velocity at t=0).

## Homework Equations

a = d2x/dt2

x(t) = 0.5at2+v0t+x0

## The Attempt at a Solution

Hi everyone, thank you for the help. I am struggling with this one as I don't exactly know how to start. After doing some research online, it seems that

d2x/dt2 is equal to vdv/dx, which can help towards solving the problem, but I don't understand why these are equal, or what was done to get there. Thank you for the help.

kuruman
Homework Helper
Gold Member
2021 Award
First off, you must assume that the acceleration is constant. Using a = dv/dt, do a first integral to find v(t). Then integrate again to find x(t).

Ross Nichols
First off, you must assume that the acceleration is constant. Using a = dv/dt, do a first integral to find v(t). Then integrate again to find x(t).

Thank you, I think I was overcomplicating the problem in my head and you were able to pull my thoughts down to earth. Here is what I got:

v = Antiderivative(a)
v(t) = at + C
When t=0, C = v0
So: v(t) = at + v0
Next, the second integral:
x(t) = antiderivative(at + v0)
x(t) = 0.5at2 +v0t + C
When t = 0, C = x0
Resulting with: x(t) = 0.5at2 +v0t + x0

Thank you for the help.