Question

Another boat in a current The water in a river moves south at \(5 \mathrm{km} / \mathrm{hr} .\) If a motorboat is traveling due east at a speed of \(40 \mathrm{km} / \mathrm{hr}\) relative to the water, determine the speed of the boat relative to the shore.

Short answer

Answer: The approximate speed of the boat relative to the shore is 40.31 km/h.

Step-by-step solution

Represent the boat and current velocities as vectors

First, we need to represent the velocities of boat and current as vectors. Let's denote the velocity of the boat relative to the water as \(\vec{v}_b\), and the velocity of the water as \(\vec{v}_w\). Since the boat is traveling east at a speed of 40 km/h, and the river current is moving south at a speed of 5 km/h, we can write the vectors as follows: \(\vec{v}_b = 40 \hat{i}\) \(\vec{v}_w = -5 \hat{j}\) Note that we used a negative sign for the river current in the south direction because it is going in the opposite direction.

Add the vectors to find the boat's velocity relative to the shore

Now that we have represented the boat and current velocities as vectors, we can add them to find the boat's velocity relative to the shore. The velocity of the boat relative to the shore, denoted as \(\vec{v}_{bs}\), can be found by: \(\vec{v}_{bs} = \vec{v}_b + \vec{v}_w\) Substituting the values from Step 1, we have: \(\vec{v}_{bs} = 40 \hat{i} - 5 \hat{j}\)

Calculate the magnitude of the boat's velocity relative to the shore

To find the speed of the boat relative to the shore, we need to calculate the magnitude of the vector \(\vec{v}_{bs}\). We can do this by using the Pythagorean theorem: \(|\vec{v}_{bs}| = \sqrt{(\vec{v}_{bs} \cdot \hat{i})^2 + (\vec{v}_{bs} \cdot \hat{j})^2}\) Plugging in the values from Step 2, we have: \(|\vec{v}_{bs}| = \sqrt{(40)^2 + (-5)^2}\) \(|\vec{v}_{bs}| = \sqrt{1600 + 25}\) \(|\vec{v}_{bs}| = \sqrt{1625}\)

Express the answer in km/h

The speed of the boat relative to the shore is the magnitude of the vector \(\vec{v}_{bs}\). Since we have calculated \(|\vec{v}_{bs}| = \sqrt{1625}\), and the units of the given speeds were in km/h, we can now find the speed of the boat relative to the shore: Speed of boat relative to shore = \(\sqrt{1625} \;\text{km/h} \approx 40.31 \;\text{km/h}\) So, the speed of the boat relative to the shore is approximately \(40.31 \;\text{km/h}\).

Key concepts

Velocity Vectors

In physics, understanding velocity vectors is essential for analyzing motion accurately. A vector is an entity that has both magnitude and direction. When we discuss velocity, a vector representation helps encompass how fast something is moving and in which direction.

For example, if a motorboat travels east at 40 km/h, its velocity vector can be represented as \( \vec{v}_b = 40 \hat{i} \), where \( \hat{i} \) denotes the eastward unit vector. The water in the river adds another dimension to this motion with its southward flow at a speed of 5 km/h. This can be expressed as a velocity vector \( \vec{v}_w = -5 \hat{j} \), where \( \hat{j} \) indicates the southward direction, and the negative sign shows the direction opposite to north.

By breaking down movements into these vectors, we can analyze how different motions interact, such as understanding the boat's overall path on the river.

Relative Motion

Relative motion considers the movement of one object with respect to another, allowing us to understand how an object appears to move from different reference points. This concept is crucial when observing an object amidst other moving bodies, like a boat moving in a flowing river.

In the context of the given problem, the motorboat's speed is relative to the water it is navigating through, but we are interested in its speed relative to the stationary shore. The river current affects the apparent motion of the boat. By using the velocity vectors discussed earlier, we combine the boat's velocity relative to the water and the river's flow to find the boat's velocity relative to the shore:

• Boat velocity = 40 km/h east
• River flow = 5 km/h south
• The combined effect: \( \vec{v}_{bs} = 40 \hat{i} - 5 \hat{j} \)

The result is the boat's actual path as observed from the stationary riverbank, taking into account both its engine-powered motion and the effect of the river's natural flow.

Pythagorean Theorem

The Pythagorean Theorem is a powerful mathematical tool that allows us to determine the magnitude of a vector when dealing with perpendicular components of motion. It states that in a right triangle, the square of the hypotenuse is the sum of the squares of the other two sides.

In vector terms, if you have a vector with two perpendicular components, \( \hat{i} \) (east) and \( \hat{j} \) (south), the magnitude of their resultant vector \( \vec{v}_{bs} \) is given by:

• \(|\vec{v}_{bs}| = \sqrt{(40)^2 + (-5)^2}\)

Calculating this, we find:

\( |\vec{v}_{bs}| = \sqrt{1600 + 25} = \sqrt{1625} \approx 40.31 \) km/h

Therefore, the boat's speed relative to the shore is about 40.31 km/h. The Pythagorean Theorem is instrumental in converting component vectors into a real-world linear velocity, providing a complete picture of the boat's motion as seen from a fixed reference point like the riverbank.