Integral of Acceleration with respect to time

  • #1

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.
 

Answers and Replies

  • #2
kuruman
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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).
 
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  • #3
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.
 

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