Question

In each of the following problems, z represents the displacement of a particle from the origin. Find (as functions of t) its speed and the magnitude of its acceleration, and describe the motion.

$\mathit{z}\mathbf{=}\left(1+i\right)\mathit{t}\mathbf{-}\left(2+i\right)\left(1-t\right)$Hint: Show that the particle moves along a straight line through the points$\mathbf{1}\mathbf{+}\mathit{i}$and$\mathbf{-}\mathbf{2}\mathbf{-}\mathit{i}$.

Short answer

The part it follows is straight line.

Its speed is, .

Acceleration

$z_i=-\left(2+i\right)\mathrm{and}{z}_f=1+i$

Step-by-step solution

Given Information.

The given equation is $z=(1+i)t-(2+i)(1-t)$.

Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$in which x is the real part and y is the imaginary part.

Simplify.

Rewrite the given equation.

$$\begin{gathered} z\left(t\right)=\left(1+i\right)t-\left(2+i\right)\left(1-t\right) \\ z\left(t\right)=\left(1+i\right)t+\left(2+i\right)t-\left(2+i\right) \\ z\left(t\right)=\left[1+i+2+i\right]t-\left(2+i\right) \\ z\left(t\right)=\left(3+2i\right)t-\left(2+i\right) \end{gathered}$$

It is of the form$z\left(t\right)=at+b$

Therefore, it represents a straight line.

Find the velocity and acceleration.

Differentiate with respect to time.

$\begin{array}{l}v\left(t\right)=\frac{dz\left(t\right)}{dt}\\ v\left(t\right\}=\frac{d}{dt}\left[\left(3+2i\right)t-(2+i)\right]\\ v\left(t\right)=3+2i\end{array}$

Find the magnitude of the velocity.

$$\begin{gathered} v\left(t\right)=\left|3+2i\right| \\ v\left(t\right)=\sqrt{3^2+2^2} \\ v\left(t\right)=\sqrt{13}\mathrm{m}/\mathrm{s} \end{gathered}$$

Find the acceleration.

Differentiate the velocity with respect to time.

$A=0$

Find the initial and final position.

Find$z\left(0\right)$ for the initial position.

$z_i=-\left(2+i\right)$

Find Final position.

$$\begin{gathered} z_f+z_i=3+2I-\left(2+I\right) \\ {Z}_f+z_i=1+i \end{gathered}$$

Therefore, the path it follow is straight line.

Its speed is $\left|v\left(t\right)=\sqrt{13}\right|$.

Acceleration 0

$z_i=-\left(2+i\right)\mathrm{and}{z}_f=1+I$