Question

Given that the acceleration vector is a(t)=(−1cos(1t))i+(−1sin(1t))j+(3t)k, the initial velocity is v(0)=i+k, and the initial position vector is r(0)=i+j+k, compute:
The Velocity Vector v(t): _ i + _ j +_k

The position vector r(t): _i + _ j + _k

Note: the coefficients in your answers must be entered in the form of expressions in the variable \emph{t};

Answer 1

Kinda hard to tell what $a(t)$ is... In any case, you can find $v(t)$ and $r(t)$ by using the fundamental theorem of calculus:

$v(t)=\displaystyle v(0)+\int_0^ta(u)\,\mathrm du$

$r(t)=\displaystyle r(0)+\int_0^tv(u)\,\mathrm du$

If you meant

$a(t)=-\cos t\,\vec\imath-\sin t\,\vec\jmath+3t\,\vec k$

then we have

$\displaystyle\int_0^t(-\cos u)\,\mathrm du=-\sin t$

$\displaystyle\int_0^t(-\sin u)\,\mathrm du=\cos t$

$\displaystyle\int_0^t3u\,\mathrm du=\frac{3t^2}2$

and so

$v(t)=(\vec\imath+\vec k)+\left(-\sin t\,\vec\imath+\cos t\,\vec\jmath+\dfrac32t^2\,\vec k\right)$

$v(t)=(1-\sin t)\,\vec\imath+\cos t\,\vec\jmath+\left(1+\dfrac{3t^2}2\right)\,\vec k$

then

$\displaystyle\int_0^t(1-\sin u)\,\mathrm du=t+\cos t$

$\displaystyle\int_0^t\cos u\,\mathrm du=\sin t$

$\displaystyle\int_0^t\frac{3u^2}2\,\mathrm du=\frac{t^3}2$

so that

$r(t)=(\vec\imath+\vec\jmath+\vec k)+\left((t+\cos t)\,\vec\imath+\sin t\,\vec\jmath+\frac{t^3}2\,\vec k\right)$

$r(t)=(1+t+\cos t)\,\vec\imath+(1+\sin t)\,\vec\jmath+\dfrac{2+t^3}2\,\vec k$