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How to Calculate Ferry Bearing to Counteract Tidal Current | Relative Velocity Physics Problem
Physics
Grade 11 of Senior High School
Question Content
A ferry is to cross the Sound of Islay from Port Askaig on Islay to Feolin on Jura which is 0.950 km due east of Port Askaig. The tidal current in the Sound of Islay is strong and the water is flowing at 3.50 m s⁻¹ in a northerly direction. The ferry travels at a speed of 5.00 m s⁻¹ relative to the water. Part A: In what direction should the ferry set out? Give your answer as a bearing.
Correct Answer
Bearing of 125° (or 180° - 55° = 125°, can also be expressed as S 55° E, but bearing format is 125°)
Detailed Solution Steps
1
Step 1: Define the required resultant velocity. The ferry needs to travel directly east relative to the land to reach Feolin, so its resultant velocity vector relative to the ground must point due east. This means the north-south component of the ferry's velocity relative to the ground must be zero.
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Step 2: Analyze velocity vectors. Let the ferry's velocity relative to water be \( \vec{v}_{fw} = 5.00 \, m/s \) at an unknown bearing, the water's velocity relative to ground be \( \vec{v}_{wg} = 3.50 \, m/s \) due north, and the ferry's velocity relative to ground be \( \vec{v}_{fg} \) due east. To cancel the northward current, the ferry must have a southward component of velocity relative to water equal to 3.50 m/s.
3
Step 3: Calculate the angle below the east direction. Use the sine function for the southward component: \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3.50}{5.00} = 0.7 \). Solve for \( \theta \): \( \theta = \arcsin(0.7) \approx 44.4° \), which is the angle south of east.
4
Step 4: Convert to bearing. Bearings are measured clockwise from north. East is 90° from north, so adding the southward angle: \( 90° + 44.4° \approx 134.4° \). A more precise calculation using \( \arcsin(0.7) = 44.427° \) gives a bearing of approximately 134°; if using cosine for the east component: \( \cos\theta = \frac{\sqrt{5^2 - 3.5^2}}{5} = \frac{\sqrt{12.75}}{5} \approx 0.714 \), \( \theta \approx 44.4° \), same result. Alternatively, some conventions calculate as 180° - 45.6° = 134.4°, rounded to 134° or 135° for simplicity.
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Note: A common precise value is bearing 134° (or 135° for rounded angle)
Knowledge Points Involved
1
Relative Velocity
Relative velocity describes the velocity of an object with respect to another reference frame, expressed by the formula \( \vec{v}_{fg} = \vec{v}_{fw} + \vec{v}_{wg} \), where \( \vec{v}_{fg} \) is velocity of ferry relative to ground, \( \vec{v}_{fw} \) is velocity of ferry relative to water, and \( \vec{v}_{wg} \) is velocity of water relative to ground. It is used to analyze motion in moving mediums like water or air.
2
Vector Components
Any vector can be broken into perpendicular components (e.g., north-south and east-west for 2D motion). For a vector of magnitude \( v \) at angle \( \theta \) to an axis, the component along the axis is \( v\cos\theta \) and perpendicular is \( v\sin\theta \). This allows solving for unknown angles or magnitudes by setting components equal to required values.
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Bearing Measurement
Bearings are angular measurements used in navigation, defined as the angle measured clockwise from the north direction. They range from 0° (due north) to 360°, providing a standardized way to describe direction relative to geographic north.
4
Trigonometric Ratios for Right Triangles
Sine (\( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \)), cosine (\( \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)) and tangent (\( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \)) relate the angles of a right triangle to its side lengths. They are used to calculate unknown angles or side lengths in vector analysis problems.
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