Question

Two ships, A and B, leave port at the same time. Ship A travels northwest at 24 knots, and ship B travels at 28 knots in a direction 40 degrees west of south. (1 knot-1 nautical mile per hour) What are the
(a) magnitude and
(b) direction of the velocity of ship A relative to B?
(c) After what time will the ships be 160 nautical miles apart?
(d) What will be the bearing of B (the direction of B's position) relative to A at that time?

Answer 1

Answer:

a) V=38.51knots

b) 88.5 degrees north of east

or 1.5 degrees east of north.

c) 4.15 hours

d) 88.5 degrees South of west

or 1.5 degrees West of South.

Explanation:

Fisrt of all, we need to find the X and Y components of the velocity of each ship.

for ship A:

$V_{ax} =V*cos(\alpha)\\V_{ax} =-24*cos(45)\\V_{ax} =-17 knots$

$V_{ay} =V*sin(\alpha)\\V_{ay =24*sin(45)\\ V_{ay} =17 knots$

Now for B:

$V_{bx} =V*sin(\alpha)\\V_{bx} =-28*sin(40)\\V_{ax} =-18 knots$

$V_{by} =V*cos(\alpha)\\V_{bx} =-28*cos(40)\\V_{ay} =-21.5 knots$

the relative velocity is given by:

$V_{ab} =V_{a} -V_{b}$

For the X axis:

$V_{abx} =(-17)-(-18)\\V_{abx} =1knot$

For the Y axis:

$V_{abx} =(17)-(-21.5)\\V_{abx} =38.5knot$

And the magnitud is given by:

$V=\sqrt{V_{abx}^2+V_{aby}^2} \\V=\sqrt{1^2+38.5^2} \\V=38.51knots$

We can calulate the angle using :

$\alpha=arctang(\frac{V_{aby}}{V_{abx}})\\\alpha=88.5^o$

that is 88.5 degrees north of east

or 1.5 degrees east of north.

In order to know the time the ships will be 160 nautical miles apart is given by:

$t=\frac{distance}{velocity}\\t=\frac{160}{38.51}=4.15h$

Because the velocity of A relative to B will be the opposite, so will be its direction.

So the angle will be:

88.5 degrees South of west

1.5 degrees West of South.