Two boats leave a marina at the same time. The first boat travels 8 knots at a bearing of , and the second boat travels 4 knots at a bearing of .
Part:
Part 1 of 2
(a) How far apart are the boats at the end of 2 hr ? Round to the nearest tenth.
The boats are approximately

Ask by Mcdonald Estrada. in the United States

Mar 25,2025

Question

Two boats leave a marina at the same time. The first boat travels 8 knots at a bearing of , and the second boat travels 4 knots at a bearing of .
Part:
Part 1 of 2
(a) How far apart are the boats at the end of 2 hr ? Round to the nearest tenth.
The boats are approximately

Ask by Mcdonald Estrada. in the United States

Mar 25,2025

UpStudy AI · Accepted Answer

Let the marina be the origin. We first find each boat’s displacement after 2 hours.

**Boat 1**

- Speed: 8 knots  
- Time: 2 hr  
- Distance:  
  $$
  d_1 = 8 \times 2 = 16 \text{ nm}
  $$
- Bearing: $$ \mathrm{N}24^\circ\mathrm{E} $$ means from north, rotate $$ 24^\circ $$ eastward. In a coordinate system with east as the $$ x $$-axis and north as the $$ y $$-axis, the components are:  
  $$
  x_1 = 16\sin(24^\circ),\quad y_1 = 16\cos(24^\circ)
  $$

**Boat 2**

- Speed: 4 knots  
- Time: 2 hr  
- Distance:  
  $$
  d_2 = 4 \times 2 = 8 \text{ nm}
  $$
- Bearing: Given as $$ 573^\circ \mathrm{W} $$. Bearings are measured modulo $$ 360^\circ $$. Thus,  
  $$
  573^\circ - 360^\circ = 213^\circ
  $$
  This means Boat 2 travels on a bearing of $$ 213^\circ $$ (measured clockwise from north). Its components are:  
  $$
  x_2 = 8\sin(213^\circ),\quad y_2 = 8\cos(213^\circ)
  $$
  
  Notice that since $$ 213^\circ = 180^\circ + 33^\circ $$:  
  $$
  \sin(213^\circ) = -\sin(33^\circ),\quad \cos(213^\circ) = -\cos(33^\circ)
  $$

**Calculating the Components**

1. **Boat 1:**
   $$
   x_1 = 16\sin(24^\circ) \approx 16(0.407) \approx 6.5 \text{ nm}
   $$
   $$
   y_1 = 16\cos(24^\circ) \approx 16(0.9135) \approx 14.6 \text{ nm}
   $$

2. **Boat 2:**
   $$
   \sin(213^\circ) \approx -\sin(33^\circ) \approx -0.544,\quad \cos(213^\circ) \approx -\cos(33^\circ) \approx -0.839
   $$
   $$
   x_2 = 8(-0.544) \approx -4.35 \text{ nm}
   $$
   $$
   y_2 = 8(-0.839) \approx -6.71 \text{ nm}
   $$

**Finding the Distance Between the Boats**

The difference in their positions is:
$$
\Delta x = x_1 - x_2 \approx 6.5 - (-4.35) = 10.85 \text{ nm}
$$
$$
\Delta y = y_1 - y_2 \approx 14.6 - (-6.71) = 21.31 \text{ nm}
$$

The distance $$ D $$ between the boats is:
$$
D = \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx \sqrt{(10.85)^2 + (21.31)^2}
$$
$$
(10.85)^2 \approx 117.7,\quad (21.31)^2 \approx 454.0
$$
$$
D \approx \sqrt{117.7 + 454.0} \approx \sqrt{571.7} \approx 23.9 \text{ nm}
$$

Thus, the boats are approximately $$ 23.9 $$ nm apart.
The boats are approximately 23.9 nautical miles apart after 2 hours.

Two boats leave a marina at the same time. The first boat travels 8 knots at a bearing of , and the second boat travels 4 knots at a bearing of .
Part:
Part 1 of 2
(a) How far apart are the boats at the end of 2 hr ? Round to the nearest tenth.
The boats are approximately

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Two boats leave a marina at the same time. The first boat travels 8 knots at a bearing of , and the second boat travels 4 knots at a bearing of . Part: Part 1 of 2 (a) How far apart are the boats at the end of 2 hr ? Round to the nearest tenth. The boats are approximately

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Two boats leave a marina at the same time. The first boat travels 8 knots at a bearing of , and the second boat travels 4 knots at a bearing of .
Part:
Part 1 of 2
(a) How far apart are the boats at the end of 2 hr ? Round to the nearest tenth.
The boats are approximately