Question

The equation r (t) = (8t + 5)i + (8t^2 - 2)j + (6t )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

Answer:

The particle's velocity is  the derivative of the particle's position. The particles's acceleration is the derivative of the particle's velocity. You can compute the velocity and acceleration as follows:

Step-by-step explanation:

$\vec{v}(t)=\frac{dr}{dt}=\frac{d}{dt}(8t+5)\overrightarrow{i}+\frac{d}{dt}(8t^2-2)\overrightarrow{j}+\frac{d}{dt}(6t)\overrightarrow{k}=8\overrightarrow{i}+16t\overrightarrow{j}+6\overrightarrow{k}.$

$\vec{a}(t)=\frac{dv}{dt}=\frac{d}{dt}(8)\overrightarrow{i}+\frac{d}{dt}(16t)\overrightarrow{j}+\frac{d}{dt}(6)\overrightarrow{k}=0\overrightarrow{i}+16\overrightarrow{j}+0\overrrighatarrow{k}.$

The velocity at $t=0$ is $\vec{v}(0)=8\overrightarrow{i}+0\overrightarrow{j}+6\overrighatarrow{k}$.

The speed at t=0 is $\lVert \vec{v}(0)\rVert =10$. Then, the velocity at t=0 written as a product of the speed at t=0 and the direction at t=0 is

$\vec{v}(0)=\lVert \vec{v}(0)\rVert \dfrac{\vec{v}(0)}{\lVert \vec{v}(0)\rVert}=10\cdot(\dfrac{8}{10},0,\dfrac{6}{10}).$