Need help with an acceleration velocity/problem. Please explain clear and concise

[DavyMathmatix]

A particle moves along the x-axis w/an acceleration given a(t) = 6t + 2, where t is measured in secs and s(position) is measured in meters. If the initial position is given s(0) = 3 and the initial velocity is given v(0)=4. Find the position of the particle at t seconds.

 

[Deleted member 4993]

A particle moves along the x-axis w/an acceleration given a(t) = 6t + 2, where t is measured in secs and s(position) is measured in meters. If the initial position is given s(0) = 3 and the initial velocity is given v(0)=4. Find the position of the particle at t seconds.

\(\displaystyle a(t) \ = \ \dfrac{d^2x}{dt^2}\)

Integrate it twice and apply the initial conditions and you are done.... cannot be any more concise!!
What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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[stapel]

A particle moves along the x-axis w/an acceleration given a(t) = 6t + 2, where t is measured in secs and s(position) is measured in meters. If the initial position is given s(0) = 3 and the initial velocity is given v(0)=4. Find the position of the particle at t seconds.

You can follow the same process that they explained and demonstrated, clearly and precisely, in your textbook:

They've given you that velocity is the integration of acceleration. How did you use this information?

When you did the integration, you got an integration constant. They've given you a data point for velocity. How did you use this information?

They've given you that position is the integration of velocity. How did you use this information?

When you did the integration, you got an integration constant. They've given you a data point for position. How did you use this information?

Please be complete, showing all of your steps. Thank you!

 

[HallsofIvy]

A particle moves along the x-axis w/an acceleration given a(t) = 6t + 2, where t is measured in secs and s(position) is measured in meters. If the initial position is given s(0) = 3 and the initial velocity is given v(0)=4. Find the position of the particle at t seconds.

The velocity function is the derivative of the position function and the acceleration function is the derivative of the velocity function. To go the other way, from acceleration to velocity and from velocity to position, do the opposite: find the anti-derivative.

Letting "v(t)" be the velocity function, you are told that \(\displaystyle a(t)= \frac{dv}{dt}= 6t+ 2\). What function has derivative 6t+ 2?

(You should know, if you are expected to do a problem like this, that \(\displaystyle \frac{d x^n}{dx}= nx^{n-1}\). Going the other way, the anti-derivative of \(\displaystyle nx^{n-1}\) is \(\displaystyle x^n\) or, replacing n-1 with m, the anti-derivative of \(\displaystyle (m+ 1)x^m\) is \(\displaystyle x^{m+1}\) so that, dividing by m+1, the anti-derivative of \(\displaystyle x^m\) is \(\displaystyle \frac{1}{m+1}x^{m+1}\). Surely you have seen that formula? Notice that, since the derivative of any constant is 0, the anti-derivative of any function can include an arbitrary added constant.)