In this scenario, a swimmer can swim at due north relative to the water, while the river current is due east. What is the swimmer’s actual velocity relative to the ground (magnitude and direction)?
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Precalculus Quiz
Practice Solving Problems With Vectors And Velocity in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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In this scenario, a swimmer can swim at vswim=4 m/s due north relative to the water, while the river current is vcurrent=3 m/s due east. What is the swimmer’s actual velocity relative to the ground (magnitude and direction)?
This quiz focuses on Solving Problems With Vectors And Velocity, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
In this scenario, a swimmer can swim at vswim=4 m/s due north relative to the water, while the river current is vcurrent=3 m/s due east. What is the swimmer’s actual velocity relative to the ground (magnitude and direction)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. Given swimmer velocity 4 m/s north and current velocity 3 m/s east, these are perpendicular, so the resultant velocity magnitude is √(4² + 3²) = √(16 + 9) = √25 = 5 m/s. The direction is arctan(3/4) = arctan(0.75) ≈ 37° east of north. Choice B is correct because it properly applies vector addition using the Pythagorean theorem for perpendicular vectors and correctly determines both magnitude (5 m/s) and direction (37° north of east, which equals 37° east of north). Choice D gives the correct magnitude but uses the wrong angle reference, stating 53° north of east when the components indicate 37° north of east (note that 53° = 90° - 37°, a common confusion between angle and its complement). Physical interpretation: if a swimmer aims north at 4 m/s but a current flows east at 3 m/s, the swimmer doesn't go straight north—they drift northeast at 5 m/s at 37° from north, which is what vector addition tells us.
For the situation described, a rescue boat's velocity relative to the water is vboat=15 km/h due west, and the ocean current is vcurrent=8 km/h due north. In what direction does the boat actually move relative to the ground (as an angle north of west)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. The rescue boat moves at 15 km/h west relative to the water, and the water itself (ocean current) moves at 8 km/h north. Since west and north are perpendicular, the resultant velocity has a northwest direction with angle arctan(northward/westward) = arctan(8/15) north of west. Choice B is correct because it properly determines the direction using arctan(8/15) north of west, where 8 is the northward component and 15 is the westward component. Choice A incorrectly uses the ratio 15/8 instead of 8/15, reversing the components in the arctangent function, which would give an angle greater than 45° when it should be less than 45° (since the westward component is larger). For perpendicular velocities or forces, the direction angle from one axis is arctan(perpendicular component/parallel component)—be careful to use the correct ratio based on which angle you're measuring from.
Given the velocities, a drone’s ground velocity is the vector sum vground=vdrone/air+vwind. The drone flies at vdrone/air=10 m/s due north, and the wind is vwind=10 m/s due south. What is the velocity of the drone relative to the ground?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. The drone flies at 10 m/s north relative to the air, and the wind blows at 10 m/s south. To find the resultant velocity relative to the ground, we add these vectors: since they're in opposite directions with equal magnitudes, we have 10 m/s north + 10 m/s south = 0 m/s. Choice B is correct because it properly applies vector addition for opposite-direction vectors of equal magnitude, correctly determining that the drone has zero velocity relative to the ground (the drone hovers in place). Choice C subtracts the velocities instead of adding them, computing v₁ - v₂ = 10 - 10 = 0 but then incorrectly assigns a direction, when actually the result is the zero vector with no direction. Physical interpretation: if a drone flies north at 10 m/s relative to the air, but the air itself moves south at 10 m/s, the drone doesn't move relative to the ground—it hovers in place, which is what vector addition tells us.
In this scenario, a drone's velocity relative to the air is vdrone=⟨12,5⟩ m/s (where +x is east and +y is north). The wind velocity is vwind=⟨−3,0⟩ m/s. What is the drone's velocity relative to the ground, vground?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition in component form. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. The drone's velocity relative to air is ⟨12, 5⟩ m/s and the wind velocity is ⟨-3, 0⟩ m/s. Adding components: v⃗_ground = ⟨12 + (-3), 5 + 0⟩ = ⟨9, 5⟩ m/s. Choice A is correct because it properly adds the velocity vectors component by component, giving ⟨9, 5⟩ m/s as the drone's velocity relative to the ground. Choice C makes an arithmetic error in the y-component, giving ⟨9, 2⟩ instead of ⟨9, 5⟩, incorrectly subtracting or modifying the northward component when the wind has no north-south component. Remember that velocity and force are vectors: when adding them in component form, simply add corresponding components—this method always works regardless of the angle between vectors.
Given the velocities, a drone’s ground velocity is the vector sum vground=vdrone+vwind. The drone’s velocity relative to the air is vdrone=⟨4,3⟩ m/s (4 m/s east, 3 m/s north). The wind is vwind=⟨−1,0⟩ m/s (1 m/s west). What is the magnitude of the resultant ground velocity ∣vground∣?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. Converting to components: drone has components ⟨4, 3⟩ and wind has components ⟨-1, 0⟩. Adding: v⃗_result = ⟨4-1, 3+0⟩ = ⟨3, 3⟩. The magnitude is √(3² + 3²) = √18 = 3√2 m/s. Choice B is correct because it properly applies vector addition using the component method. Choice A makes an arithmetic error, calculating √(5² + 3²) = √34 instead of using the correct summed components. Key to relative velocity problems: identify the object's velocity relative to the medium (drone to air) and the medium's velocity relative to the ground, then add these two vectors using components or Pythagorean theorem for perpendicular cases. Remember that velocity is a vector: when adding them, you must account for both magnitude and direction, using either the component method (always works) or geometric methods (for special cases like perpendicular vectors).
A boat travels at 12 m/s due north relative to the water, while the river current flows at 5 m/s due east relative to the ground. Given the velocities, what is the boat's actual velocity relative to the ground (magnitude and direction)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. Given boat at 12 m/s north and current at 5 m/s east, these are perpendicular, so the resultant velocity magnitude is √(12² + 5²) = √(144 + 25) = √169 = 13 m/s. The direction is arctan(5/12) ≈ 22.6° east of north. Choice A is correct because it properly applies vector addition and correctly determines magnitude and direction. Choice D uses arctan(12/5) ≈ 67.4° east of north, which reverses the ratio and gives the direction from the east axis instead of north. Key to relative velocity problems: identify the object's velocity relative to the medium (boat to water, plane to air) and the medium's velocity relative to the ground, then add these two vectors using components or Pythagorean theorem for perpendicular cases. For perpendicular velocities or forces, use the Pythagorean theorem for magnitude: |v⃗| = √(v₁² + v₂²), and arctan(v₂/v₁) for direction—this is faster than the component method when vectors are perpendicular.
For the situation described, a ball is thrown with initial velocity magnitude 20 m/s at an angle of 30∘ above the horizontal. What is the horizontal component of the velocity, vx?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector components. To find the resultant of vectors at any angle, convert each to components using vₓ = v·cos(θ) and vᵧ = v·sin(θ), add the components, then find the magnitude √(vₓ² + vᵧ²) and direction arctan(vᵧ/vₓ) of the sum. For a ball thrown at initial velocity magnitude 20 m/s at an angle of 30° above the horizontal, the horizontal component is vₓ = v·cos(θ) = 20·cos(30°) = 20·(√3/2) = 10√3 m/s. Choice B is correct because it properly applies the cosine function to find the horizontal component, giving vₓ = 20cos(30°) = 10√3 m/s. Choice C uses sine when should use cosine for the horizontal component calculation, computing 20sin(30°) = 10 m/s instead of 20cos(30°) = 10√3 m/s—sine gives the vertical component, not horizontal. Remember that velocity and force are vectors: when finding components, use cosine for the component along the reference axis (horizontal) and sine for the perpendicular component (vertical).
A ship can travel at 10 km/h relative to the water. A current flows at 6 km/h due south. The ship must end up traveling due east relative to the ground. For the situation described, what heading should the ship take (as an angle north of east)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. To achieve due east ground velocity, the ship's heading must produce a northward component to cancel the 6 km/h southward current: 10 sinθ = 6, so sinθ = 0.6, θ = arcsin(0.6) ≈ 36.9° north of east. Choice A is correct because it correctly uses sine for the component calculation to counteract the current. Choice B uses cosine when should use sine for the component calculation, computing cos^{-1}(0.6) ≈ 53.1° instead of sin^{-1}(0.6). Key to relative velocity problems: identify the object's velocity relative to the medium (boat to water, plane to air) and the medium's velocity relative to the ground, then add these two vectors using components or Pythagorean theorem for perpendicular cases. To verify your answer, check that the resultant magnitude is between |v₁ - v₂| and |v₁ + v₂| (triangle inequality), and that the direction makes physical sense given the original vectors' directions.
For the situation described, a drone’s velocity relative to the air is vdrone=20 m/s at 30∘ above the horizontal (measured from the +x axis). What is the horizontal component of the velocity, vx?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. To find the resultant of vectors at any angle, convert each to components using vₓ = v·cos(θ) and vᵧ = v·sin(θ), add the components, then find the magnitude √(vₓ² + vᵧ²) and direction arctan(vᵧ/vₓ) of the sum. Converting to components: the drone's velocity of 20 m/s at 30° has vₓ = 20 cos(30°) = 20 (√3/2) = 10√3 m/s and vᵧ = 20 sin(30°) = 20 (1/2) = 10 m/s. Choice B is correct because it correctly uses cosine for the horizontal component. Choice A uses sine when should use cosine for the component calculation, computing 20 sin(30°) instead of 20 cos(30°). Remember that velocity and force are vectors: when adding them, you must account for both magnitude and direction, using either the component method (always works) or geometric methods (for special cases like perpendicular vectors). To verify your answer, check that the resultant magnitude is between |v₁ - v₂| and |v₁ + v₂| (triangle inequality), and that the direction makes physical sense given the original vectors' directions.
A hiker's displacement is the vector sum of two displacements: first 5 km due east, then 12 km due north. For the situation described, what is the magnitude of the resultant displacement?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically displacement problems using vector addition. When two velocities or forces are perpendicular, the magnitude of the resultant can be found using the Pythagorean theorem: |v⃗_result| = √(|v⃗₁|² + |v⃗₂|²), and the direction is found using arctan(v₂/v₁). Given displacements of 5 km east and 12 km north, these are perpendicular, so the resultant displacement magnitude is √(5² + 12²) = √(25 + 144) = √169 = 13 km. Choice C is correct because it correctly uses the Pythagorean theorem for perpendicular vectors. Choice B incorrectly adds the magnitudes directly (5 + 12 = 17), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add speeds. For perpendicular velocities or forces, use the Pythagorean theorem for magnitude: |v⃗| = √(v₁² + v₂²), and arctan(v₂/v₁) for direction—this is faster than the component method when vectors are perpendicular. Remember that velocity and force are vectors: when adding them, you must account for both magnitude and direction, using either the component method (always works) or geometric methods (for special cases like perpendicular vectors).
Given the velocities, a person walks at vperson/train=2 m/s due west relative to a train, while the train moves at vtrain/ground=10 m/s due east relative to the ground. How fast is the person moving relative to the ground (magnitude and direction)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. The person walks at 2 m/s west relative to the train, and the train moves at 10 m/s east relative to the ground. To find the resultant velocity relative to the ground, we add these vectors: since they're in opposite directions, we have 10 m/s east + 2 m/s west = 10 m/s east - 2 m/s east = 8 m/s east. Choice B is correct because it properly applies vector addition for opposite-direction vectors, correctly determining that the person moves 8 m/s due east relative to the ground. Choice A incorrectly adds the magnitudes directly (10 + 2 = 12), ignoring that the velocities are in opposite directions—when vectors point opposite ways, their magnitudes subtract, not add. Key to relative velocity problems: identify the object's velocity relative to the medium (person to train) and the medium's velocity relative to the ground (train to ground), then add these two vectors using components or direct subtraction for opposite cases.
A boat travels from point A to point B, a distance of 12 km in a direction 25° east of north. The trip takes exactly 45 minutes. On the return trip from B to A, the boat encounters a current and takes 60 minutes. Assuming the boat maintains the same speed relative to the water on both trips, what is the component of the current velocity parallel to the line AB?
Explanation: Let vb be the boat's speed in still water and vc be the current component along AB. For trip A→B: ground speed = 12 km ÷ 0.75 h = 16 km/h, so vb + vc = 16. For trip B→A: ground speed = 12 km ÷ 1 h = 12 km/h, so vb - vc = 12. Solving the system: Adding equations gives 2vb = 28, so vb = 14 km/h. Substituting back: vc = 16 - 14 = 2 km/h.
In this scenario, a boat has a velocity relative to the water of vboat=12 km/h due north, and the river current has a velocity vcurrent=5 km/h due east. Given that vground=vboat+vcurrent, what is the boat's actual velocity relative to the ground (magnitude and direction)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. Given the boat velocity 12 km/h north and current velocity 5 km/h east, these are perpendicular, so the resultant velocity magnitude is √(12² + 5²) = √(144 + 25) = √169 = 13 km/h. The direction is arctan(5/12) ≈ 22.6° east of north. Choice B is correct because it properly applies vector addition using the Pythagorean theorem for perpendicular vectors and correctly determines both magnitude (13 km/h) and direction (tan⁻¹(5/12) east of north). Choice A incorrectly adds the magnitudes directly (12 + 5 = 17), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add speeds. For perpendicular velocities or forces, use the Pythagorean theorem for magnitude: |v⃗| = √(v₁² + v₂²), and arctan(v₂/v₁) for direction—this is faster than the component method when vectors are perpendicular.
For the situation described, a boat’s velocity relative to the water is vboat=4 m/s due west, and the current is vcurrent=3 m/s due north. In what direction does the boat actually move relative to the ground (angle measured north of west)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. When two velocities or forces are perpendicular, the magnitude of the resultant can be found using the Pythagorean theorem: |v⃗_result| = √(|v⃗₁|² + |v⃗₂|²), and the direction is found using arctan(v₂/v₁). The boat moves at 4 m/s west relative to the water, and the current flows at 3 m/s north. To find the resultant velocity relative to the ground, we add these vectors: the resultant has magnitude √(4² + 3²) = 5 m/s and direction arctan(3/4) ≈ 37° from the westward direction. Since the current pushes north, the boat moves northwest at 37° north of west. Choice A is correct because it properly determines the direction using arctan(3/4) = 37° and correctly identifies that the boat moves north of west (not south) due to the northward current. Choice B gives the complement angle 53° (since 53° = 90° - 37°), confusing the angle with its complement—the correct angle from west is 37°, not 53°. Physical interpretation: if a boat aims west at 4 m/s but a current flows north at 3 m/s, the boat doesn't go straight west—it drifts northwest at about 5 m/s, which is what vector addition tells us.
In this scenario, two forces act on a crate: F1=20 N east and F2=15 N north. What are the magnitude and direction of the net force Fnet?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically force problems using vector addition. Force, like velocity, is a vector quantity, and the net force on an object is the vector sum of all individual forces acting on it: F⃗_net = ΣF⃗ᵢ, found using the same vector addition methods as velocity problems. Two forces of 20 N east and 15 N north act on the object; since they are perpendicular, we use Pythagorean theorem: magnitude √(20² + 15²) = √(400 + 225) = √625 = 25 N in direction arctan(15/20) ≈ 37° north of east. Choice B is correct because it properly applies vector addition and correctly uses Pythagorean theorem for perpendicular vectors to find magnitude and direction. Choice A incorrectly adds the magnitudes directly (20 + 15 = 35), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add magnitudes. For perpendicular velocities or forces, use the Pythagorean theorem for magnitude: |v⃗| = √(v₁² + v₂²), and arctan(v₂/v₁) for direction—this is faster than the component method when vectors are perpendicular. Remember that velocity and force are vectors: when adding them, you must account for both magnitude and direction, using either the component method (always works) or geometric methods (for special cases like perpendicular vectors).
For the situation described, a hiker walks 3 km east, then 4 km north. What is the magnitude of the resultant displacement from the starting point?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically displacement problems using vector addition. To find the resultant of vectors at any angle, convert each to components using vₓ = v·cos(θ) and vᵧ = v·sin(θ), add the components, then find the magnitude √(vₓ² + vᵧ²) and direction arctan(vᵧ/vₓ) of the sum. Given displacements of 3 km east and 4 km north, these are perpendicular, so the resultant displacement magnitude is √(3² + 4²) = √(9 + 16) = √25 = 5 km. Choice C is correct because it correctly uses the Pythagorean theorem for perpendicular vectors. Choice B incorrectly adds the magnitudes directly (3 + 4 = 7), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add distances. For perpendicular velocities or forces, use the Pythagorean theorem for magnitude: |v⃗| = √(v₁² + v₂²), and arctan(v₂/v₁) for direction—this is faster than the component method when vectors are perpendicular. Physical interpretation: the straight-line displacement from start to end is the vector sum, shorter than the path traveled, illustrating the difference between distance and displacement.
Given the velocities, a hiker walks at vhiker=4 m/s due east, then encounters a moving walkway that carries people at vwalkway=3 m/s due north. Assuming the hiker continues walking due east on the walkway, what is the magnitude of the resultant velocity ∣vtotal∣ relative to the ground?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. The hiker walks at 4 m/s east relative to the walkway, and the walkway itself moves at 3 m/s north. Since these velocities are perpendicular, we use the Pythagorean theorem: |v⃗_result| = √(4² + 3²) = √(16 + 9) = √25 = 5 m/s. Choice C is correct because it properly applies the Pythagorean theorem for perpendicular vectors, recognizing that 4² + 3² = 25, so √25 = 5 m/s. Choice A incorrectly adds the magnitudes directly (4 + 3 = 7), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add speeds. Physical interpretation: the hiker moves diagonally (northeast) relative to the ground, with the walkway carrying them north while they walk east, resulting in a 5 m/s diagonal velocity.
For the situation described, a ball is thrown with initial velocity magnitude 20 m/s at 30∘ above the horizontal. What is the horizontal component of the velocity (ignore air resistance)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. To find the resultant of vectors at any angle, convert each to components using vₓ = v·cos(θ) and vᵧ = v·sin(θ), add the components, then find the magnitude √(vₓ² + vᵧ²) and direction arctan(vᵧ/vₓ) of the sum. Converting to components: initial velocity of 20 m/s at 30° has components ⟨20 cos(30°), 20 sin(30°)⟩ = ⟨10√3, 10⟩ m/s, so the horizontal component is 10√3 m/s. Choice C is correct because it correctly uses cosine for the horizontal component calculation. Choice B uses sine when should use cosine (or vice versa) for the component calculation, computing the vertical component instead of horizontal. Remember that velocity and force are vectors: when adding them, you must account for both magnitude and direction, using either the component method (always works) or geometric methods (for special cases like perpendicular vectors). Physical interpretation: the horizontal component remains constant in projectile motion (ignoring air resistance), determining the range, while the vertical changes due to gravity.
A rescue helicopter's ground velocity is given by vground=vheli+vwind. The helicopter flies at 40 m/s due north relative to the air, and the wind blows at 30 m/s due east. Given the velocities, in what direction does the helicopter actually move relative to the ground (angle east of north)?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. Velocity is a vector quantity with both magnitude (speed) and direction, and when an object moves through a moving medium (like a boat in a current or plane in wind), the resultant velocity relative to the ground is the vector sum: v⃗_resultant = v⃗_object + v⃗_medium. Given helicopter at 40 m/s north and wind at 30 m/s east, these are perpendicular, so the direction is arctan(30/40) ≈ 36.9° east of north. Choice B is correct because it properly applies vector addition and correctly determines the direction using arctan of the ratio. Choice A uses arctan(40/30) ≈ 53.1° east of north, reversing the ratio and giving the direction from the east axis instead of north. Key to relative velocity problems: identify the object's velocity relative to the medium (boat to water, plane to air) and the medium's velocity relative to the ground, then add these two vectors using components or Pythagorean theorem for perpendicular cases. For perpendicular velocities or forces, use the Pythagorean theorem for magnitude: |v⃗| = √(v₁² + v₂²), and arctan(v₂/v₁) for direction—this is faster than the component method when vectors are perpendicular.
For the situation described, an airplane’s velocity relative to the air is vplane=300 km/h due east, and the wind velocity is vwind=50 km/h due south. What is the magnitude of the plane’s ground velocity ∣vground∣?
Explanation: This question tests understanding of solving real-world problems involving vectors, specifically velocity problems using vector addition. When two velocities or forces are perpendicular, the magnitude of the resultant can be found using the Pythagorean theorem: |v⃗_result| = √(|v⃗₁|² + |v⃗₂|²), and the direction is found using arctan(v₂/v₁). The airplane moves at 300 km/h east relative to the air, and the wind itself moves at 50 km/h south. To find the resultant velocity relative to the ground, we add these vectors: since they're perpendicular, the magnitude is √(300² + 50²) km/h. Choice C is correct because it properly applies the Pythagorean theorem for perpendicular vectors, giving the exact expression √(300² + 50²) for the ground speed magnitude. Choice A incorrectly adds the magnitudes directly (300 + 50 = 350), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add speeds. Remember that velocity and force are vectors: when adding them, you must account for both magnitude and direction, using either the component method (always works) or geometric methods (for special cases like perpendicular vectors).