Problem 79 - Fundamentals of Physics

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 \(40^{\circ}\) west of south. \((1 \mathrm{knot}=1\) nautical mile per hour; see Appendix D.) 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?

Step-by-Step Solution

Answer

(a) 21.94 knots (b) 290° from east (c) 7.30 hours (d) 250° from north

Step 1: Represent Velocities as Vectors

Convert the velocities of Ships A and B into vector form. Ship A is traveling northwest, which translates to a direction of 135° from the east (since NW is halfway between north and west). Ship A's velocity vector can be written as:\[v_A = 24 \text{ knots } \begin{pmatrix} \cos(135^{\circ}) \ \sin(135^{\circ}) \end{pmatrix} = 24 \begin{pmatrix} -\frac{\sqrt{2}}{2} \ \frac{\sqrt{2}}{2} \end{pmatrix}\]Ship B travels 40° west of south. From north (0°), this places Ship B's direction at 230° (180° + 40°). Ship B's velocity vector can be written as:\[v_B = 28 \text{ knots } \begin{pmatrix} \cos(230^{\circ}) \ \sin(230^{\circ}) \end{pmatrix} = 28 \begin{pmatrix} \cos(230^{\circ}) \ \sin(230^{\circ}) \end{pmatrix}\]

Step 2: Find the Relative Velocity of A with Respect to B

The relative velocity \(v_{A,B}\) of Ship A with respect to Ship B is the vector difference between their velocity vectors:\[v_{A,B} = v_A - v_B\]Substitute the expressions for \(v_A\) and \(v_B\):\[v_{A,B} = \begin{pmatrix} 24 \cos(135^{\circ}) - 28 \cos(230^{\circ}) \ 24 \sin(135^{\circ}) - 28 \sin(230^{\circ}) \end{pmatrix}\]Calculate each component to get:\[v_{A,B} = \begin{pmatrix} -7.55 \ 20.70 \end{pmatrix}\]

Step 3: Calculate the Magnitude of the Relative Velocity

The magnitude of the relative velocity \(|v_{A,B}|\) can be found using the Pythagorean theorem:\[|v_{A,B}| = \sqrt{(-7.55)^2 + (20.70)^2} = 21.94 \text{ knots}\]

Step 4: Determine the Direction of the Relative Velocity

The direction \(\theta\) of the relative velocity vector is given by:\[\theta = \tan^{-1}\left(\frac{20.70}{-7.55}\right)\]Calculate \(\theta\):\[\theta \approx \tan^{-1}\left(-2.74\right) \approx 110.0^{\circ} \text{ from east}\]Convert this to a bearing direction: \(290.0^{\circ}\) means \(70.0^{\circ}\) west of south.

Step 5: Determine Time for Ships to be 160 Nautical Miles Apart

The distance apart is given by the relative speed multiplied by time. Set up the equation:\[160 = 21.94 t\]Solve for \(t\):\[t \approx \frac{160}{21.94} \approx 7.30 \text{ hours}\]

Step 6: Calculate the Bearing of B Relative to A After 7.30 Hours

Using the relative velocity components, after 7.30 hours, B's coordinates relative to A's starting position would be:\[(-7.55 \times 7.30, 20.70 \times 7.30) = (-55.115, 151.11)\]The bearing relative to A is:\[\text{bearing}=\tan^{-1}\left(\frac{151.11}{-55.115}\right) \approx 110.0^{\circ}\]Converting to standard bearing yields a direction of \(250^{\circ}\) from north (bearing is taken in the clockwise direction).

Key Concepts

Vector Addition

When dealing with problems involving multiple velocities, such as two ships traveling in different directions, vector addition becomes crucial. Each velocity can be represented as a vector, showing both magnitude and direction. In this context, we convert the speeds of the ships into vectors using trigonometric functions:

• Ship A, moving northwest, is directed at 135° from a positive east direction.
• Ship B, going 40° west of south, corresponds to 230° from north.

Splitting each speed into horizontal and vertical components allows us to accurately represent their motions. Calculating the relative velocity involves finding the difference between these velocity vectors, which is easily achieved by subtracting each corresponding component of the vectors.

Pythagorean Theorem

In the world of vector operations, the Pythagorean Theorem helps calculate the magnitude of a vector resulting from vector additions or subtractions. Given a resultant vector with components \(x\) and \(y\), its magnitude is calculated as:\[|\mathbf{v}| = \sqrt{x^2 + y^2}\]In our ship exercise, after computing the relative velocity vector's components (\[\begin{pmatrix} -7.55, 20.70 \end{pmatrix}\]), apply the Pythagorean Theorem to find its magnitude. This represents the speed at which one ship moves relative to the other. Thus, it provides a single value detailing how quickly two ships drift apart considering their individual paths and speeds.

Bearing Direction

Bearing is a method used to describe the direction of travel concerning north, adjusting clockwise. It is crucial for navigation. In the ship exercise, we explore the bearing of Ship B relative to Ship A. Initially, using the relative velocity vector components, we compute the angle using:\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]The calculated angle is then adjusted to derive a 'bearing', providing shipping routes in navigational terms. Here, the result of 290° roughly translates to 70° west of south, which helps navigators align with recognized directional terms rather than pure angular measures.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a vital role in dissecting vectors into manageable directions and magnitudes. In navigation scenarios:

• Cosine helps find the horizontal component: \(\cos(\theta)\).
• Sine determines the vertical component: \(\sin(\theta)\).
• The inverse tangent function, \(\tan^{-1}\), retrieves direction from given components.

In our problem involving Ships A and B, we utilize these functions to break down their velocity vectors based on polar angles. Calculation results enable accurate direction and speed evaluation, thereby facilitating the ships' relative movement assessment. Understanding these functions is key, transforming navigational challenges into straightforward decomposed tasks.