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

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:

Explanation:

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

`\vec{a}(t)=(dv)/(dt)=(d)/(dt)(8)\overrightarrow{i}+(d)/(dt)(16t)\overrightarrow{j}+(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 \frac{\vec{v}(0)}{\lVert \vec{v}(0)\rVert}=10\cdot((8)/(10),0,(6)/(10)).`