Question

Find the position vector R(t) and velocity vector V(t), given the acceleration A(t) and initial position and velocity vectors R(0) and V(0), respectively.
A(t)= t^2i-2t^(1/2)j+e^(3t)k; R(0)= 2i+j-k; V(0)=i-j-2k

Answer 1

The fundamental theorem of calculus tells us that

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

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

So we have

$\vec v(t)=(\vec\imath-\vec\jmath-2\,\vec k)+\displaystyle\int_0^t(u^2\,\vec\imath-2u^{1/2}\,\vec\jmath+e^{3u}\,\vec k)\,\mathrm du$

$\vec v(t)=(\vec\imath-\vec\jmath-2\,\vec k)+\left(\dfrac{u^3}3\,\vec\imath-\dfrac{4u^{3/2}}3\,\vec\jmath+\dfrac{e^{3u}}3\,\vec k\right)\bigg|_0^t$

$\vec v(t)=\dfrac{t^3+3}3\,\vec\imath-\dfrac{4t^{3/2}+3}3\,\vec\jmath+\dfrac{e^{3t}-7}3\,\vec k$

and

$\vec r(t)=(2\,\vec\imath+\vec\jmath-\vec k)+\displaystyle\int_0^t\left(\frac{u^3+3}3\,\vec\imath-\frac{4u^{3/2}+3}3\,\vec\jmath+\frac{e^{3u}-7}3\right)\,\mathrm du$

$\vec r(t)=(2\,\vec\imath+\vec\jmath-\vec k)+\left(\dfrac{u^3+12u}{12}\,\vec\imath-\dfrac{8u^{5/2}+15u}{15}\,\vec\jmath+\dfrac{e^{3u}-21u}9\,\vec k\right)\bigg|_0^t$

$\vec r(t)=\dfrac{t^3+12t+24}{12}\,\vec\imath-\dfrac{8t^{5/2}+15t-15}{15}\,\vec\jmath+\dfrac{e^{3t}-21t-10}9\,\vec k$