Problem 90 - Physics for Scientists and Engineers with Modern Physics

Question

A boat can travel \(2.20 \mathrm{~m} / \mathrm{s}\) in still water. \((a)\) If the boat points its prow directly across a stream whose current is \(1.30 \mathrm{~m} / \mathrm{s},\) what is the velocity (magnitude and direction) of the boat relative to the shore? \((b)\) What will be the position of the boat, relative to its point of origin, after \(3.00 \mathrm{~s}\) ?

Step-by-Step Solution

Answer

The boat travels at approximately 2.56 m/s at an angle of 30.8° to the shore, reaching a position of (6.60 m, 3.90 m) after 3 seconds.

Step 1: Understand the Problem

We need to find the velocity of the boat relative to the shore when it's crossing a stream with a current. The boat's velocity in still water is given, as well as the current's velocity. We're also asked to find the boat's position after 3 seconds.

Step 2: Use Vector Addition to Find Velocity

The velocity of the boat relative to the water is 2.20 m/s perpendicular to the shore (across the stream), and the stream's current velocity is 1.30 m/s parallel to the shoreline. Use vector addition to find the resultant velocity of the boat relative to the shore. The components of velocity are: \( v_{bx} = 2.20 \text{ m/s} \) and \( v_{by} = 1.30 \text{ m/s} \).

Step 3: Calculate the Magnitude of Velocity

The magnitude of the boat's velocity relative to the shore can be calculated using the Pythagorean theorem: \[ v = \sqrt{v_{bx}^2 + v_{by}^2} = \sqrt{(2.20)^2 + (1.30)^2} \] Evaluate this to find the magnitude.

Step 4: Find the Direction of Velocity

The direction of the velocity relative to the shore can be found using the tangent function: \[ \theta = \tan^{-1}\left(\frac{v_{by}}{v_{bx}}\right) = \tan^{-1}\left(\frac{1.30}{2.20}\right) \] Calculate \(\theta\) to find the direction angle with respect to the direction directly across the stream.

Step 5: Calculate the Position After 3 Seconds

To find the position of the boat after 3 seconds, use the component velocities. The displacement in the x-direction is \( d_x = v_{bx} \times t = 2.20 \times 3 \), and in the y-direction, it's \( d_y = v_{by} \times t = 1.30 \times 3 \). These provide the resultant position vector.

Key Concepts

Vector Addition

When trying to determine the resulting effect of multiple velocities acting together, we employ the concept of vector addition. In this scenario, the problem involves a boat moving across a stream where two vector quantities are in play: the boat's own velocity in still water and the stream's current velocity.

To compute the net or resultant velocity of the boat relative to the shore, each velocity is considered a vector. The boat's velocity across the stream is 2.20 m/s, acting perpendicular to the shoreline. Meanwhile, the stream flows parallel to the shore with a velocity of 1.30 m/s.

To find the resulting velocity, you can

• Draw and label each velocity as a vector originating from a common point.
• Apply the vector addition principle, which combines these vectors into a resultant vector or final velocity of the boat relative to the shore.

This method of vector addition ensures all directions and magnitudes are properly accounted for, so you can accurately gauge the boat's movement from the shore's perspective.

Pythagorean Theorem

Once you have identified the individual velocity components, the Pythagorean theorem comes into play to calculate the magnitude of the resultant vector.

Since vector components of velocities create a right triangle (with perpendicular sides) when depicted, you can directly apply this well-known theorem. It states that in a right triangle, the square of the hypotenuse (resultant vector) is equal to the sum of the squares of the other two sides (individual velocity vectors).

In formulaic terms, if we denote the sides as \( v_{bx} = 2.20 \text{ m/s} \) and \( v_{by} = 1.30 \text{ m/s} \), the hypotenuse, or resultant velocity \( v \), is determined as follows:

• Use the equation: \[ v = \sqrt{v_{bx}^2 + v_{by}^2} = \sqrt{(2.20)^2 + (1.30)^2} \]
• Evaluate to find the magnitude of the resultant vector.

This outcome specifies how fast and at what angle the boat moves across the stream from the surrounding land's point of view.

Displacement Calculation

The next goal is to determine how far the boat will have moved from its starting position after a specified duration, namely 3 seconds. This involves calculating the boat's displacement using its calculated velocity components.

Displacement is essentially determining the change in position of the boat over the time interval. For this, you can apply the basic principles of linear motion to each velocity component:

• The horizontal displacement \( d_x \) is given by the formula: \( d_x = v_{bx} \times t = 2.20 \times 3 \).
• Similarly, the vertical displacement \( d_y \) is calculated by: \( d_y = v_{by} \times t = 1.30 \times 3 \).

The resultant position vector, combining these two displacements, tells you exactly where the boat is located relative to its initial position, taking into account both the boat's motion and the stream's impact. Understanding displacement in this way helps gauge the journey's progress over time seamlessly.