Problem 81 - Fundamentals of Physics

Question

\(A\) is located \(4.0 \mathrm{~km}\) north and \(2.5 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(22 \mathrm{~km} / \mathrm{h}\) toward the south, and ship \(B\) has a velocity of \(40 \mathrm{~km} / \mathrm{h}\) in a direction \(37^{\circ}\) north of east. (a) What is the velocity of \(A\) relative to \(B\) in unit-vector notation with \(\hat{\mathrm{i}}\) toward the east? (b) Write an expression (in terms of \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) ) for the position of \(A\) relative to \(B\) as a function of \(t\), where \(t=0\) when the ships are in the positions described above. (c) At what time is the separation between the ships least? (d) What is that least separation?

Step-by-Step Solution

Answer

(a) \(-{40\cos(37^\circ)}\hat{\mathbf{i}} - (22 + 40\sin(37^\circ))\hat{\mathbf{j}}\) km/h. (b) \((2.5 - 40\cos(37^\circ)t)\hat{\mathbf{i}} + (4.0 - (22 + 40\sin(37^\circ))t)\hat{\mathbf{j}}\) km. (c) Solve derivative for minimum. (d) Calculate least distance from (c).

Step 1: Set up unit vector notation for velocities

First, we convert the velocities of ships A and B into unit-vector notation. Ship A moves south, so its velocity vector is \(-22\hat{\mathbf{j}}\) km/h (since south corresponds to the negative \(\hat{\mathbf{j}}\) direction). Ship B's velocity needs to be broken into components because it moves 37° north of east. Its eastward component is \(40\cos(37^\circ)\) and the northward component is \(40\sin(37^\circ)\). Thus, the velocity vector for B is \(\mathbf{v}_B = 40\cos(37^\circ)\hat{\mathbf{i}} + 40\sin(37^\circ)\hat{\mathbf{j}}\).

Step 2: Find the velocity of A relative to B

To find the velocity of A relative to B, we subtract B's velocity from A's velocity. Using the vectors established: \(\mathbf{v}_{\text{relative}} = \mathbf{v}_A - \mathbf{v}_B = -22\hat{\mathbf{j}} - (40\cos(37^\circ)\hat{\mathbf{i}} + 40\sin(37^\circ)\hat{\mathbf{j}})\). Simplify to get: \(-40\cos(37^\circ)\hat{\mathbf{i}} - (22 + 40\sin(37^\circ))\hat{\mathbf{j}}\).

Step 3: Determine initial relative position vector

Initially, ship A is 4.0 km north and 2.5 km east of ship B. This means the initial relative position vector from B to A is \(\mathbf{r}_{0} = 2.5\hat{\mathbf{i}} + 4.0\hat{\mathbf{j}}\).

Step 4: Write expression for position of A relative to B as a function of time

The position of A relative to B as a function of time is given by their initial position plus the relative velocity multiplied by time: \(\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_{\text{relative}} t\). Substituting the earlier results gives: \(\mathbf{r}(t) = (2.5 - 40\cos(37^\circ) t)\hat{\mathbf{i}} + (4.0 - (22 + 40\sin(37^\circ)) t)\hat{\mathbf{j}}\).

Step 5: Find time of least separation by setting the derivative of distance to zero

To find when the separation is least, compute the magnitude of the position vector as a function of time, \(d(t) = \sqrt{(2.5 - 40\cos(37^\circ) t)^2 + (4.0 - (22 + 40\sin(37^\circ)) t)^2}\), and set its derivative to zero. Analytically solve this (or use a computational tool) to find the critical time \(t_0\).

Step 6: Calculate least separation

Once \(t_0\) is found, substitute it into the position function \(\mathbf{r}(t)\) to find \(\mathbf{r}(t_0)\). Compute the magnitude \(d(t_0)\) to get the least separation.

Key Concepts

Unit-Vector Notation

When analyzing motion, especially in two or three dimensions, unit-vector notation provides a clear way to express direction and magnitude together. A unit vector is a vector with a magnitude of one and indicates the direction along a particular axis. In our scenario:

• The direction to the east is represented by the unit vector \(\hat{\mathbf{i}}\), aligned with the x-axis.
• The direction to the north is represented by the unit vector \(\hat{\mathbf{j}}\), aligned with the y-axis.

This framework helps simplify the calculations of relative motion between two ships moving along specified directions. By denoting velocities and positions using unit-vector notation, we can easily break down complex movements into manageable components along these axes.

Velocity Components

Velocity is a vector quantity, which means it has both magnitude and direction. When a ship moves in a direction that is not aligned with one of the coordinate axes, we need to decompose its velocity into components.
For ship B, it moves 37° north of east, so:

• The eastward (or horizontal) component of ship B’s velocity is calculated using \(40\cos(37^\circ)\).
• The northward (or vertical) component is \(40\sin(37^\circ)\).

Ship A's velocity is simpler because it moves directly southward, resulting in a single negative component along the \(\hat{\mathbf{j}}\) direction.
Using these components, we encapsulate the ship's velocities in a mathematical form that's conducive to further calculations like finding relative velocity.

Differentiation

Differentiation is a critical concept in calculus used to find rates of change. In our exercise, we employ differentiation to find the rate of change of distance between the ships.
By expressing the position of one ship relative to another as a function of time, we can determine how this distance changes over time.

• The function for distance is expressed in terms of time \(t\), embedding velocity and initial positions.
• Differentiating this function with respect to \(t\) helps identify when the rate of change is zero, signaling the point of minimum separation.

Understanding how to apply differentiation to these functions allows us to solve for the critical times when key changes, like minimum separation, occur.

Relative Position

Relative position pinpoints the location of one object concerning another. It's useful for tracking the dynamic relationship between moving entities, like ships.
Initially, ship A is located 4.0 km north and 2.5 km east of ship B. This setup is neatly expressed as a relative position vector: \(2.5\hat{\mathbf{i}} + 4.0\hat{\mathbf{j}}\).
By integrating the relative velocity over time, we create an expression for position as a function of time:

• The relative position changes as both ships move, affected by their respective velocities.
• With the expression \(\mathbf{r}(t) = (2.5 - 40\cos(37^\circ) t)\hat{\mathbf{i}} + (4.0 - (22 + 40\sin(37^\circ)) t)\hat{\mathbf{j}}\), you can plug in various time instances to find positions dynamically.

This approach is invaluable in predicting future configurations, allowing navigators to avoid potential collisions or maintain desired proximities.