Math Problem Statement

  1. A sailboat heads north at 3m/s for 1 hour and then back to the southeast (at 45 to north) at 2m/s for 45 minutes. (2 points) A) How far has the boat sailed? B) How far is it from its starting location?

Solution

Let's break down the problem into two parts to find the required answers.

Part A: Total Distance Sailed

  1. Northward Journey:

    • Speed: 3 m/s
    • Time: 1 hour (3600 seconds)
    • Distance d1d_1: d1=speed×time=3m/s×3600s=10800md_1 = \text{speed} \times \text{time} = 3 \, \text{m/s} \times 3600 \, \text{s} = 10800 \, \text{m}
  2. Southeastward Journey:

    • Speed: 2 m/s
    • Time: 45 minutes (2700 seconds)
    • Distance d2d_2: d2=speed×time=2m/s×2700s=5400md_2 = \text{speed} \times \text{time} = 2 \, \text{m/s} \times 2700 \, \text{s} = 5400 \, \text{m}

Total Distance Sailed:

Total Distance=d1+d2=10800m+5400m=16200m\text{Total Distance} = d_1 + d_2 = 10800 \, \text{m} + 5400 \, \text{m} = 16200 \, \text{m}

Part B: Displacement from Starting Point

  1. Northward Displacement:

    • The boat travels 10800 m directly north.
  2. Southeastward Displacement:

    • The southeast direction is 45° to the north, so it can be broken down into components:
      • Northward component d2Nd_{2N}: d2N=d2×cos(45)=5400×123818.5md_{2N} = d_2 \times \cos(45^\circ) = 5400 \times \frac{1}{\sqrt{2}} \approx 3818.5 \, \text{m}
      • Eastward component d2Ed_{2E}: d2E=d2×sin(45)=5400×123818.5md_{2E} = d_2 \times \sin(45^\circ) = 5400 \times \frac{1}{\sqrt{2}} \approx 3818.5 \, \text{m}
  3. Net Displacement in Northward Direction: dN=108003818.56981.5md_N = 10800 - 3818.5 \approx 6981.5 \, \text{m}

  4. Net Displacement in Eastward Direction: dE=3818.5md_E = 3818.5 \, \text{m}

  5. Magnitude of Displacement: Displacement=dN2+dE2=6981.52+3818.5248740482.25+14581292.2563321774.57957.5m\text{Displacement} = \sqrt{d_N^2 + d_E^2} = \sqrt{6981.5^2 + 3818.5^2} \approx \sqrt{48740482.25 + 14581292.25} \approx \sqrt{63321774.5} \approx 7957.5 \, \text{m}

Final Answers:

A) The total distance sailed by the boat is 16,200 meters.

B) The boat is approximately 7,957.5 meters away from its starting location.

Would you like further details or have any questions? Here are some related questions:

  1. What is the difference between total distance traveled and displacement?
  2. How do angles affect the direction of travel in vector problems?
  3. What are some examples of breaking down vectors into components?
  4. How can you use trigonometry to find the components of a vector?
  5. What are other real-life examples where vector addition is important?

Tip: Always draw a diagram when dealing with vector problems. It helps in visualizing the problem and breaking down the components accurately.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Vector Addition
Displacement
Distance

Formulas

Distance = Speed × Time
Vector components: Vx = V × cos(θ), Vy = V × sin(θ)
Pythagorean Theorem: c = √(a^2 + b^2)

Theorems

Pythagorean Theorem
Trigonometric Identities

Suitable Grade Level

Grades 9-12