Question

Find the position vector when acceleration a(t)= 2i + j + 3k

Answer 1

Speed: $v(t)=\int{a(t)dt}=\int{(2i+j+3k)}dt=2ti+tj+3tk+C.$

Position: $s(t)=\int{v(t)dt}=\int{(2ti+tj+3yk+C)dt}=t^2i+\frac{1}{2}t^2j+\frac{3}{2}t^2k+Ct+d.$