A ship leaves port at 8 knots (nautical miles per hour) heading . After 2 hr , it makes a clockwise turn to a new bearing of and travels for
1.5 hr .
Part 1 of 2

Ask by Turner Osborne. in the United States

Mar 13,2025

Question

A ship leaves port at 8 knots (nautical miles per hour) heading . After 2 hr , it makes a clockwise turn to a new bearing of and travels for
1.5 hr .
Part 1 of 2

Ask by Turner Osborne. in the United States

Mar 13,2025

UpStudy AI · Accepted Answer

We first resolve each leg into north–south and east–west components.

---

1. **First Leg (2 hr at 8 knots on a bearing of $$ \mathrm{N}29^\circ \mathrm{W} $$)**

• The distance traveled is
$$
16\text{ nm} = 8 \times 2.
$$

• The bearing $$ \mathrm{N}29^\circ \mathrm{W} $$ means the course is $$29^\circ$$ west of due north. Its components are:
  - North component: 
$$
16\cos(29^\circ).
$$
  - West component (negative in the east–west coordinate if east is positive):
$$
-16\sin(29^\circ).
$$

Thus, after the first leg the position relative to port is
$$
(x_1,y_1)=\Bigl(-16\sin(29^\circ),\,16\cos(29^\circ)\Bigr).
$$

---

2. **Second Leg (1.5 hr at 8 knots on a bearing of $$ \mathrm{N}61^\circ \mathrm{E} $$)**

• The distance traveled is
$$
12\text{ nm} = 8 \times 1.5.
$$

• The bearing $$ \mathrm{N}61^\circ \mathrm{E} $$ means the course is $$61^\circ$$ east of due north. Its components are:
  - North component: 
$$
12\cos(61^\circ).
$$
  - East component:
$$
12\sin(61^\circ).
$$

Thus, the displacement for the second leg is
$$
(x_2,y_2)=\Bigl(12\sin(61^\circ),\,12\cos(61^\circ)\Bigr).
$$

---

3. **Final Position Relative to Port**

We add the components of both legs:
$$
\begin{aligned}
x &= x_1+x_2 = -16\sin(29^\circ)+12\sin(61^\circ),\[1mm]
y &= y_1+y_2 = 16\cos(29^\circ)+12\cos(61^\circ).
\end{aligned}
$$

Noticing that numerically
$$
\sin(29^\circ)\approx0.4841,\quad \cos(29^\circ)\approx0.8746,
$$
$$
\sin(61^\circ)\approx0.8746,\quad \cos(61^\circ)\approx0.4841,
$$
we have
$$
\begin{aligned}
x&\approx -16(0.4841)+12(0.8746)\[1mm]
&\approx -7.746+10.495\[1mm]
&\approx 2.75~\text{nm},\[2mm]
y&\approx 16(0.8746)+12(0.4841)\[1mm]
&\approx 13.994+5.809\[1mm]
&\approx 19.80~\text{nm}.
\end{aligned}
$$

Thus, the ship is approximately $$2.75$$ nautical miles east and $$19.80$$ nautical miles north of the port.
After traveling 2 hours at 8 knots on a bearing of $$ \mathrm{N}29^\circ \mathrm{W} $$, the ship is approximately 2.75 nautical miles east and 19.80 nautical miles north of the port. Then, after turning $$90^\circ$$ clockwise to a bearing of $$ \mathrm{N}61^\circ \mathrm{E} $$ and traveling 1.5 hours at 8 knots, the ship's final position is approximately 2.75 nautical miles east and 19.80 nautical miles north of the port.

A ship leaves port at 8 knots (nautical miles per hour) heading . After 2 hr , it makes a clockwise turn to a new bearing of and travels for
1.5 hr .
Part 1 of 2

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Trigonometry Questions

A ship leaves port at 8 knots (nautical miles per hour) heading . After 2 hr , it makes a clockwise turn to a new bearing of and travels for 1.5 hr . Part 1 of 2

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Trigonometry Questions

A ship leaves port at 8 knots (nautical miles per hour) heading . After 2 hr , it makes a clockwise turn to a new bearing of and travels for
1.5 hr .
Part 1 of 2