# Thread: maximum velocity of a particle with 3d vector equation

1. ## maximum velocity of a particle with 3d vector equation

Find the maximum speed of a particle whose velocity v m/s, at any time t seconds is given by v$\displaystyle =2sin(t)$ i $\displaystyle + cos(t)$ j $\displaystyle + 3$ k.

2. Originally Posted by shawli
Find the maximum speed of a particle whose velocity v m/s, at any time t seconds is given by v$\displaystyle =2sin(t)$ i $\displaystyle + cos(t)$ j $\displaystyle + 3$ k.
Dear shawli,

$\displaystyle v=2\sin{t}\underline{i}+\cos{t}\underline{j}+3\und erline{k}$

$\displaystyle Therefore,~speed~of~the~particle~(S)=\left|2\sin{t }\underline{i}+\cos{t}\underline{j}+3\underline{k} \right|$

$\displaystyle S=\sqrt{(2sint)^2+cos^{2}t+9}$

$\displaystyle S=\sqrt{3sin^{2}t+10}$

$\displaystyle Since,~(sin^{2}t)_{max}=1$

$\displaystyle S_{max}=\sqrt{13}\approx{3.61}$

3. $\displaystyle \|\vec{v}\|^2=(2sin(t))^2+(cos(t))^2+9=3sin^2t+10\ leq 13 ~,~ \|\vec{v}\|_{max}=\sqrt{13}$