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First time posting, so please advise if and where I'm braking protocols. Here is the question I was given:

A particle travels with acceleration given by $a(t)=(2e^{-t})i+(5\cos{t})j-(3\sin{t})k$. When the particle is located at $(1,-3,2)$ at time $t=0$ and is moving with a velocity given by $v(t)=4i-3j+2k$, find the velocity and displacement of the particle at any time $t>0$.

I know the relation between velocity and acceleration is $a(t)=dv/dt$, and displacement is the integral of $v(t)$ across an interval. But I'm not sure how to set this one up. Wouldn't the velocity just be the integral of acceleration? And then the displacement is the integral of the velocity? But why then was I given the position and velocity at $t=0$?

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closed as off-topic by Daniel Griscom, Sebastian Riese, user36790, John Rennie, Kyle Kanos Jan 24 '16 at 10:52

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Yes, you are right that velocity is the integral of acceleration, and displacement is in turn the integral of velocity. The reason that you were given those points is that when you integrate, you get a +C constant. You can find the value of that constant by plugging in those points for velocity and position at time t=0.

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Please do the integrations component wise . Then you will get the components of velocity and displacement by integrating the components of acceleration consecutively. And I think you know how to find those integration constants from the boundary conditions

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