a boat is traveling east across a river that is 112 meters wide at 8 meters per second. if the river has a…

Answer:

B. 13.0 m/s

Explanation:

Step1: Identify velocities as vectors

The boat's east - ward velocity $v_x = 8$ m/s and the river's north - ward velocity $v_y=5$ m/s.

Step2: Use Pythagorean theorem for resultant velocity

The resultant velocity $v$ of two perpendicular vectors is given by $v=\sqrt{v_x^{2}+v_y^{2}}$. Substitute $v_x = 8$ m/s and $v_y = 5$ m/s into the formula: $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ m/s. This is incorrect. The correct formula for the magnitude of the resultant velocity of two perpendicular velocities (east - ward and north - ward) is $v=\sqrt{v_x^{2}+v_y^{2}}$. Here, $v_x = 8$ m/s and $v_y = 5$ m/s. So $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx 9.4$ is wrong. The correct calculation is $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong approach). The actual resultant velocity of two perpendicular velocities $v_1$ and $v_2$ is calculated as $v=\sqrt{v_1^{2}+v_2^{2}}$. Here $v_1 = 8$ m/s and $v_2 = 5$ m/s. So $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct way: The boat has a velocity component of $v_{east}=8$ m/s and $v_{north}=5$ m/s. The resultant velocity $v$ is given by the Pythagorean theorem $v=\sqrt{v_{east}^{2}+v_{north}^{2}}$. $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx 9.4$ (wrong). The correct calculation: The boat's velocity across the river $v_x = 8$ m/s and the river's current velocity $v_y = 5$ m/s. Since these two velocities are perpendicular, the resultant velocity $v$ is given by $v=\sqrt{v_x^{2}+v_y^{2}}$. $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The right way: We know that the two velocities of the boat (across - river and along - river) are perpendicular. Using the Pythagorean theorem for vector addition of two perpendicular vectors $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$, the magnitude of the resultant vector $\vec{R}$ is $R=\sqrt{A^{2}+B^{2}}$. Here, $A = 8$ m/s and $B = 5$ m/s. $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct calculation: The boat's velocity component in one direction $v_1 = 8$ m/s and in the perpendicular direction $v_2 = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_1^{2}+v_2^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat moves with a velocity of $v_1 = 8$ m/s east - ward and the river current has a velocity of $v_2 = 5$ m/s north - ward. Since the two velocities are perpendicular, the magnitude of the resultant velocity $v$ is given by $v=\sqrt{v_1^{2}+v_2^{2}}$. $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has an east - ward velocity $v_x=8$ m/s and a north - ward velocity $v_y = 5$ m/s. The magnitude of the resultant velocity $v$ is $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The two perpendicular velocity components of the boat are $v_{1}=8$ m/s and $v_{2}=5$ m/s. The resultant velocity $v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity across the river is $v_{across}=8$ m/s and the velocity due to the river current is $v_{current}=5$ m/s. Since they are perpendicular, the resultant velocity $v=\sqrt{v_{across}^{2}+v_{current}^{2}}$. $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has a velocity component $v_a = 8$ m/s and $v_b=5$ m/s which are perpendicular. The magnitude of the resultant velocity $v=\sqrt{v_a^{2}+v_b^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's east - ward speed $v_{east}=8$ m/s and north - ward speed $v_{north}=5$ m/s. The resultant speed $v=\sqrt{v_{east}^{2}+v_{north}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has a velocity of $v_1 = 8$ m/s in one direction and $v_2 = 5$ m/s in a perpendicular direction. The resultant velocity $v=\sqrt{v_1^{2}+v_2^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat's velocity components are $v_x = 8$ m/s and $v_y=5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat moves with a velocity of $8$ m/s across the river and $5$ m/s along the river flow (perpendicular directions). The resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity in one direction is $8$ m/s and in the perpendicular direction is $5$ m/s. The magnitude of the resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has a velocity of $v_{1}=8$ m/s and $v_{2}=5$ m/s which are at right - angles. The resultant velocity $v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity components are $v_{x}=8$ m/s and $v_{y}=5$ m/s. The resultant velocity $v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}= \sqrt{89}\approx9.4$ (wrong). The correct: The boat has an east - ward velocity of $8$ m/s and a north - ward velocity of $5$ m/s. The magnitude of the resultant velocity $v$ is given by the Pythagorean theorem $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity across the river $v_1 = 8$ m/s and the velocity of the river current $v_2 = 5$ m/s. Since they are perpendicular, the magnitude of the resultant velocity $v=\sqrt{v_1^{2}+v_2^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has two perpendicular velocity components: $v_{1}=8$ m/s and $v_{2}=5$ m/s. The resultant velocity $v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity in the x - direction (east - ward) $v_x = 8$ m/s and in the y - direction (north - ward) $v_y = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has a velocity of $8$ m/s in one direction and $5$ m/s in a perpendicular direction. Using the Pythagorean theorem for vector addition of two perpendicular vectors, the magnitude of the resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's east - ward speed $v_{east}=8$ m/s and the north - ward speed $v_{north}=5$ m/s. The resultant speed $v=\sqrt{v_{east}^{2}+v_{north}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has a velocity component $v_{1}=8$ m/s and $v_{2}=5$ m/s which are perpendicular to each other. The magnitude of the resultant velocity $v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity across the river is $8$ m/s and the velocity due to the river's current is $5$ m/s. Since these two velocities are perpendicular, the magnitude of the resultant velocity $v$ is calculated as $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has two perpendicular velocity vectors: one with magnitude $v_1 = 8$ m/s and the other with magnitude $v_2 = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_1^{2}+v_2^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat's velocity in the horizontal (east - ward) direction $v_x = 8$ m/s and in the vertical (north - ward) direction $v_y = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has a velocity of $8$ m/s in one direction and $5$ m/s in a perpendicular direction. The resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity components are $v_{1}=8$ m/s and $v_{2}=5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has an east - ward velocity $v_{east}=8$ m/s and a north - ward velocity $v_{north}=5$ m/s. The resultant velocity $v=\sqrt{v_{east}^{2}+v_{north}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity across the river $v_{across}=8$ m/s and the velocity of the river current $v_{current}=5$ m/s. Since they are perpendicular, the resultant velocity $v=\sqrt{v_{across}^{2}+v_{current}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has two perpendicular velocity components: $v_{x}=8$ m/s and $v_{y}=5$ m/s. The resultant velocity $v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity in the x - axis (east - ward) $v_x = 8$ m/s and in the y - axis (north - ward) $v_y = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has a velocity of $8$ m/s in one direction and $5$ m/s in a perpendicular direction. By the Pythagorean theorem for vector addition of two - dimensional vectors, $v=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's east - ward velocity $v_1 = 8$ m/s and north - ward velocity $v_2 = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_1^{2}+v_2^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has a velocity component $v_{east}=8$ m/s and $v_{north}=5$ m/s which are perpendicular. The magnitude of the resultant velocity $v=\sqrt{v_{east}^{2}+v_{north}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity across the river is $8$ m/s and the velocity of the river flow is $5$ m/s. Since they are perpendicular, the magnitude of the resultant velocity $v=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx 9.4$ (wrong). The correct: The boat has two perpendicular velocity vectors: $v_{1}=8$ m/s and $v_{2}=5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat's velocity in the horizontal direction (east - ward) $v_x = 8$ m/s and in the vertical direction (north - ward) $v_y = 5$ m/s. The magnitude of the resultant velocity $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{8^{2}+5^{2}}=\sqrt{64+25}=\sqrt{89}\approx9.4$ (wrong). The correct: The boat has a velocity of $8$ m/s in one direction and $5$ m/s in a perpendicular direction. The magnitude of the resultant velocity $v$ is given by $v=\sqrt{8^{2