Hi please help me with this using MATLAB code, thank you!

a$)$ Complete the script that generates a table showing the position $(x,y)$

of each ship as a function of time for the next three $(3)$ hours.

b$)$ Expand the output table to display the distance between the two ships

as a function of time.

c$)$ If the visibility is $10$ miles, when $[$HH:MM$],$ and for how long

$($minutes$),$ would people from the two ships be able to see each other?

Extra$-$credit $[15$ points$]$

Plot the distance $($d$)$ between the two ships as a function of time $($ t $)$

$d_{A-B}=\sqrt[\phantom{2}]{(x_A-x_B{)}^2+(y_A-y_B{)}^2}$

Note: to receive full credit your script must use Tables and

Lists

The fundamental equations are

$x=x^0+V_{x}t+\frac{1}{2}{a}_x{t}^2$

$y=y^0+V_{y}t+\frac{1}{2}{a}_y{t}^2$

Here the ships travel at a constant speed $($i$.$e$.$ the accelerations are $a_x=$

$a_y=0!)$Two ships, A and B$,$ depart from two different ports. Ship $"$A$"$ travels

towards the NE $($i$.$e$.,$ towards the East in an angle, $,$ of $30$ North$)$ at a

constant $(a_x=a_y=0)$ speed of $14$ miles per hour, while Ship $"B"$ travels

South at a constant $(a_x=a_y=0)$ speed of $7$ miles per hour. At $7$:$30$ AM$,$

the ships are positioned as shown in the schematic below:

At $7$:$30$ AM

Ship $"$A$"$

$x_A^o=-s_{A}cos$

$y_A^0=-s_{A}sin$

$V_{A,x}=V_{A}cos$

$V_{A,y}=V_{A}sin$

Ship $"$B$"$

$x_B^o=0$

$y_B^o=s_B$

$V_{B,x}=0$

$V_{B,y}=-V_B$