Integral of Acceleration with respect to time

[Ross Nichols]

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]

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.