Question

The equation $r\left(t\right)=(8t+5)i+(8t^2-2)j+\left(6t\right)k$ is the position of a particle in space at time $t$. Find the​ particle's velocity and acceleration vectors. Then write the​ particle's velocity at $t$ equals $0$ as a product of its speed and direction. What is the velocity​ vector?

Answer 1

Find the velocity vector of the particle in space-time $\mathit{t}$.

The vector equation of the particle is,

$r\left(t\right)=(8t+5)i+(8t^2-2)j+\left(6t\right)k$

Step 1. Calculate the velocity of the particle.

$\begin{array}{rcl}v\left(t\right)& =& r\text{'}\left(t\right)\\ & \Rightarrow & v\left(t\right)=\frac{d}{dt}\left((8t+5)i+(8t^2-2)j+\left(6t\right)k\right)\\ & \Rightarrow & v\left(t\right)=\left(8+0\right)i+\left(8\left(2t\right)-0\right)j+6k\\ & \Rightarrow & v\left(t\right)=8i+16tj+6k\end{array}$

Step 2. Calculate the acceleration of the particle.

$\begin{array}{rcl}a\left(t\right)& =& r\text{'}\text{'}\left(t\right)\\ & \Rightarrow & a\left(t\right)=\frac{d}{dt}\left(r\text{'}\left(t\right)\right)\\ & \Rightarrow & a\left(t\right)=\frac{d}{dt}\left(8i+16tj+6k\right)\\ & \Rightarrow & a\left(t\right)=0+16j+0\\ & \Rightarrow & a\left(t\right)=+16j\end{array}$

Step 3. Calculate the speed of the particle at $\mathit{t}\mathbf{=}\mathbf{0}$.

$\begin{array}{rcl}\left|v\left(t\right)\right|& =& \left|8i+16tj+6k\right|\\ & \Rightarrow & \left|v\left(t\right)\right|=\sqrt{8^2+{\left(16t\right)}^2+{\left(6\right)}^2}\\ & \Rightarrow & \left|v\left(t\right)\right|=\sqrt{64+256t^2+36}\\ & \Rightarrow & \left|v\left(t\right)\right|=\sqrt{256t^2+100}\\ & \Rightarrow & \left|v\left(0\right)\right|=\sqrt{256{\left(0\right)}^2+100}\\ & \Rightarrow & \left|v\left(0\right)\right|=\sqrt{100}\\ & \Rightarrow & \left|v\left(0\right)\right|=10\end{array}$

Step 4. Calculate the direction of the particle at $\mathit{t}\mathbf{=}\mathbf{0}$.

$\begin{array}{rcl}\frac{v\left(t\right)}{\left|v\left(t\right)\right|}& =& \frac{8i+16tj+6k}{10}\\ & \Rightarrow & \frac{v\left(0\right)}{\left|v\left(0\right)\right|}=\frac{8i+16\left(0\right)j+6k}{10}\\ & \Rightarrow & \frac{v\left(0\right)}{\left|v\left(0\right)\right|}=\frac{8i+6k}{10}\\ & \Rightarrow & \frac{v\left(0\right)}{\left|v\left(0\right)\right|}=\frac{4}{5}i+\frac{3}{5}k\\ & \Rightarrow & \frac{v\left(0\right)}{\left|v\left(0\right)\right|}=\frac{1}{5}\left(4i+3k\right)\end{array}$

Since the​ particle's velocity at $t$ equals $0$ as a product of its speed 10 and direction $\frac{1}{5}\left(4i+3k\right)$.

Therefore, the velocity vector is $\mathbf{10}\left(\frac{4}{5}i+\frac{3}{5}k\right)$.