Position Problem

[Jason76]

The acceleration at time \(\displaystyle t\), of particle moving along the \(\displaystyle x\) axis is given by \(\displaystyle a(t) = 30t^{3} + 6\) At time \(\displaystyle = 0\), the velocity of the particle is \(\displaystyle 0\) and the position of the particle is \(\displaystyle 6\). What is the position of the particle at time \(\displaystyle t\)?

Any starting hints?

 

[MarkFL]

The time rate of change of velocity is acceleration, thus:

\(\displaystyle \dfrac{dv}{dt}\equiv a(t)\)

So, what could you do to find \(\displaystyle v(t)\)?

Then, use the fact that the time rate of change of position is velocity:

\(\displaystyle \dfrac{dx}{dt}\equiv v(t)\)

So, what could you do to find \(\displaystyle x(t)\)?

 

[Jason76]

The time rate of change of velocity is acceleration, thus:

\(\displaystyle \dfrac{dv}{dt}\equiv a(t)\)

So, what could you do to find \(\displaystyle v(t)\)?

Then, use the fact that the time rate of change of position is velocity:

\(\displaystyle \dfrac{dx}{dt}\equiv v(t)\)

So, what could you do to find \(\displaystyle x(t)\)?

y' of position is velocity, and the y' of velocity is acceleration. Is that right? So, if the equation given is the acceleration, then taking the integral should give the velocity, and then taking integral of that, should give the position.

 

[Deleted member 4993]

Rate of change of position with respect to time is velocity, and the Rate of change of velocity with respect to time is acceleration. Is that right? So, if the equation given is the acceleration, then taking the integral should give the velocity, and then taking integral of that, should give the position.

Correct - except for the corrections shown.