Instantaneous acceleration/ velocity

  • Thread starter lylek
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Hi everyone, first time poster.

I was just hoping somebody could explain instantaneous acceleration to me a bit better than my prof did. I know that we take dv / dt. But I am just having trouble applying it. Does this mean that I am putting the derivative x's position equation on top? and say we are looking at t = 3s, what does d (3) mean.

Thanks for any help.
 

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  • #2
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Does this mean that I am putting the derivative x's position equation on top?
What do you mean with that?

If you know the x-position as function of time (x(t)), you can calculate the derivative v(t) - the velocity.
Now, you can calculate the derivative of v(t), and this is the acceleration a(t).
 
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HallsofIvy
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Hi everyone, first time poster.

I was just hoping somebody could explain instantaneous acceleration to me a bit better than my prof did. I know that we take dv / dt. But I am just having trouble applying it. Does this mean that I am putting the derivative x's position equation on top? and say we are looking at t = 3s, what does d (3) mean.

Thanks for any help.
If you are using x(t) to mean the position of the object at time t, then the velocity is given by v= dx/dt and the acceleration is a= dv/dt= d^2x/dt^2, the second derivative of x with respect to t. If, by "position on top" you mean that you have trouble remembering if v= dx/dt or dt/dx, it might help you to remember that velocity , v, is "meters per second" or m/s and acceleration, a, is "meters per second squared" or m/s^2 so the time is always in the denominator.
 

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