Precalculus Quiz: Solving Problems With Vectors And Velocity

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Solving Problems With Vectors And VelocityQuestion 1 of 20

In this scenario, a swimmer can swim at $\vec v_{\text{swim}}=4\ \text{m/s}$ due north relative to the water, while the river current is $\vec v_{\text{current}}=3\ \text{m/s}$ due east. What is the swimmer's actual velocity relative to the ground (magnitude and direction)?

1\ \text{m/s}

37^\circ

north of east

5\ \text{m/s}

37^\circ

north of east

7\ \text{m/s}

53^\circ

north of east

5\ \text{m/s}

53^\circ

north of east

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Precalculus Quiz

Precalculus Quiz: Solving Problems With Vectors And Velocity

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.

What this quiz covers

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.

How to use this quiz

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.

All questions

Question 1

$1\ \text{m/s}$ at $37^\circ$ north of east
$5\ \text{m/s}$ at $37^\circ$ north of east (correct answer)
$7\ \text{m/s}$ at $53^\circ$ north of east
$5\ \text{m/s}$ at $53^\circ$ 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. 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.

Question 2

For the situation described, a rescue boat's velocity relative to the water is $\vec v_{\text{boat}}=15\text{ km/h}$ due west, and the ocean current is $\vec v_{\text{current}}=8\text{ km/h}$ due north. In what direction does the boat actually move relative to the ground (as an angle north of west)?

$\tan^{-1}\!\left(\frac{15}{8}\right)$ north of west
$\tan^{-1}\!\left(\frac{8}{15}\right)$ north of west (correct answer)
$\tan^{-1}\!\left(\frac{8}{15}\right)$ south of west
$\tan^{-1}\!\left(\frac{15}{8}\right)$ south 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.

Question 3

Given the velocities, a drone's ground velocity is the vector sum $\vec v_{\text{ground}}=\vec v_{\text{drone/air}}+\vec v_{\text{wind}}$ . The drone flies at $\vec v_{\text{drone/air}}=10\ \text{m/s}$ due north, and the wind is $\vec v_{\text{wind}}=10\ \text{m/s}$ due south. What is the velocity of the drone relative to the ground?

$20\ \text{m/s}$ due north
$0\ \text{m/s}$ (the zero vector) (correct answer)
$10\ \text{m/s}$ due north
$10\ \text{m/s}$ due south

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.

Question 4

In this scenario, a drone's velocity relative to the air is $\vec v_{\text{drone}}=\langle 12,\,5\rangle\text{ m/s}$ (where $+x$ is east and $+y$ is north). The wind velocity is $\vec v_{\text{wind}}=\langle -3,\,0\rangle\text{ m/s}$ . What is the drone's velocity relative to the ground, $\vec v_{\text{ground}}$ ?

$\langle 9,\,5\rangle\text{ m/s}$ (correct answer)
$\langle 15,\,5\rangle\text{ m/s}$
$\langle 9,\,2\rangle\text{ m/s}$
$\langle -9,\,-5\rangle\text{ m/s}$

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.

Question 5

Given the velocities, a drone's ground velocity is the vector sum $\vec v_{ground}=\vec v_{drone}+\vec v_{wind}$ . The drone's velocity relative to the air is $\vec v_{drone}=\langle 4,3\rangle$ m/s (4 m/s east, 3 m/s north). The wind is $\vec v_{wind}=\langle -1,0\rangle$ m/s (1 m/s west). What is the magnitude of the resultant ground velocity $|\vec v_{ground}|$ ?

$\sqrt{(5)^2+(3)^2}=\sqrt{34}$ m/s
$\sqrt{(3)^2+(3)^2}=3\sqrt{2}$ m/s (correct answer)
$\sqrt{(4)^2+(3)^2}=5$ m/s
$\sqrt{(4-3)^2+(1)^2}=\sqrt{2}$ m/s

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).

Question 6

A boat travels at $12\ \text{m/s}$ due north relative to the water, while the river current flows at $5\ \text{m/s}$ due east relative to the ground. Given the velocities, what is the boat's actual velocity relative to the ground (magnitude and direction)?

$13\ \text{m/s}$ at $\tan^{-1}(5/12)\approx 22.6^\circ$ east of north (correct answer)
$17\ \text{m/s}$ due northeast
$7\ \text{m/s}$ at $22.6^\circ$ west of north
$13\ \text{m/s}$ at $\tan^{-1}(12/5)\approx 67.4^\circ$ 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 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.

Question 7

A ship can travel at $10\ \text{km/h}$ relative to the water. A current flows at $6\ \text{km/h}$ 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)?

$\sin^{-1}(6/10)\approx 36.9^\circ$ north of east (correct answer)
$\cos^{-1}(6/10)\approx 53.1^\circ$ north of east
$\tan^{-1}(6/10)\approx 31.0^\circ$ north of east
$\sin^{-1}(10/6)\approx 59.0^\circ$ 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.

Question 8

For the situation described, a drone's velocity relative to the air is $\vec v_{\text{drone}}=20\ \text{m/s}$ at $30^\circ$ above the horizontal (measured from the +x axis). What is the horizontal component of the velocity, $v_x$ ?

$20\sin(30^\circ)=10\ \text{m/s}$
$20\cos(30^\circ)=10\sqrt{3}\ \text{m/s}$ (correct answer)
$20\cos(30^\circ)=10\ \text{m/s}$
$20\sin(30^\circ)=10\sqrt{3}\ \text{m/s}$

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.

Question 9

A hiker's displacement is the vector sum of two displacements: first $5\ \text{km}$ due east, then $12\ \text{km}$ due north. For the situation described, what is the magnitude of the resultant displacement?

$7\ \text{km}$
$17\ \text{km}$
$13\ \text{km}$ (correct answer)
$\sqrt{119}\ \text{km}$

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).

Question 10

Given the velocities, a person walks at $\vec v_{\text{person/train}}=2\ \text{m/s}$ due west relative to a train, while the train moves at $\vec v_{\text{train/ground}}=10\ \text{m/s}$ due east relative to the ground. How fast is the person moving relative to the ground (magnitude and direction)?

$12\ \text{m/s}$ due east
$8\ \text{m/s}$ due east (correct answer)
$8\ \text{m/s}$ due west
$10\ \text{m/s}$ due 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. 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.

Question 11

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?

1.2 km/h
1.6 km/h
2.0 km/h (correct answer)
2.4 km/h

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.

Question 12

In this scenario, a boat has a velocity relative to the water of $\vec v_{\text{boat}}=12\ \text{km/h}$ due north, and the river current has a velocity $\vec v_{\text{current}}=5\ \text{km/h}$ due east. Given that $\vec v_{\text{ground}}=\vec v_{\text{boat}}+\vec v_{\text{current}}$ , what is the boat's actual velocity relative to the ground (magnitude and direction)?

$17\ \text{km/h}$ due northeast
$13\ \text{km/h}$ at $\tan^{-1}(5/12)$ east of north (correct answer)
$7\ \text{km/h}$ at $\tan^{-1}(5/12)$ east of north
$13\ \text{km/h}$ at $\tan^{-1}(12/5)$ 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 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.

Question 13

For the situation described, a boat's velocity relative to the water is $\vec v_{\text{boat}}=4\ \text{m/s}$ due west, and the current is $\vec v_{\text{current}}=3\ \text{m/s}$ due north. In what direction does the boat actually move relative to the ground (angle measured north of west)?

$37^\circ$ north of west (correct answer)
$53^\circ$ north of west
$37^\circ$ south of west
$53^\circ$ south 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.

Question 14

Given the velocities, a hiker walks at $\vec v_{\text{hiker}}=4\text{ m/s}$ due east, then encounters a moving walkway that carries people at $\vec v_{\text{walkway}}=3\text{ m/s}$ due north. Assuming the hiker continues walking due east on the walkway, what is the magnitude of the resultant velocity $|\vec v_{\text{total}}|$ relative to the ground?

$7\text{ m/s}$
$1\text{ m/s}$
$5\text{ m/s}$ (correct answer)
$\sqrt{4^2-3^2}\text{ m/s}$

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.

Question 15

A rescue helicopter's ground velocity is given by $\vec v_{\text{ground}}=\vec v_{\text{heli}}+\vec v_{\text{wind}}$ . The helicopter flies at $40\ \text{m/s}$ due north relative to the air, and the wind blows at $30\ \text{m/s}$ due east. Given the velocities, in what direction does the helicopter actually move relative to the ground (angle east of north)?

$\tan^{-1}(40/30)\approx 53.1^\circ$ east of north
$\tan^{-1}(30/40)\approx 36.9^\circ$ east of north (correct answer)
$\tan^{-1}(30/40)\approx 36.9^\circ$ west of north
Due 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.

Question 16

For the situation described, an airplane's velocity relative to the air is $\vec v_{\text{plane}}=300\ \text{km/h}$ due east, and the wind velocity is $\vec v_{\text{wind}}=50\ \text{km/h}$ due south. What is the magnitude of the plane's ground velocity $|\vec v_{\text{ground}}|$ ?

$350\ \text{km/h}$
$250\ \text{km/h}$
$\sqrt{300^2+50^2}\ \text{km/h}$ (correct answer)
$\sqrt{300^2-50^2}\ \text{km/h}$

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).

Question 17

In this scenario, a ball is thrown with initial velocity magnitude $20\text{ m/s}$ at $30^\circ$ above the horizontal. What is the horizontal component of the velocity, $v_x$ ?

$20\sin(30^\circ)\text{ m/s}$
$20\cos(30^\circ)\text{ m/s}$ (correct answer)
$20\tan(30^\circ)\text{ m/s}$
$\frac{20}{\cos(30^\circ)}\text{ m/s}$

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 velocity of 20 m/s at 30° above the horizontal, the horizontal component is vₓ = 20·cos(30°) m/s. Choice B is correct because it properly uses cosine to find the horizontal component of a vector given at an angle above the horizontal. Choice A incorrectly uses sine when should use cosine for the horizontal component, computing 20·sin(30°) instead of 20·cos(30°)—sine gives the vertical component, not horizontal. Remember that for a vector at angle θ from the horizontal: horizontal component uses cosine (vₓ = v·cos(θ)) and vertical component uses sine (vᵧ = v·sin(θ))—this is fundamental to decomposing vectors into components.

Question 18

Given the velocities, an airplane has airspeed $300$ km/h due east, and the wind blows $50$ km/h due south. What is the magnitude of the resultant ground velocity $|\vec v_{ground}|$ ?

$350$ km/h
$250$ km/h
$\sqrt{92500}$ km/h (correct answer)
$\sqrt{65000}$ km/h

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 airplane at 300 km/h east and wind at 50 km/h south, these are perpendicular, so the resultant velocity magnitude is √(300² + 50²) = √(90000 + 2500) = √92500 km/h. Choice C is correct because it correctly uses the Pythagorean theorem for perpendicular vectors. 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. 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. 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.

Question 19

In this scenario, two forces act on a crate on a frictionless floor: $\vec F_1=20$ N east and $\vec F_2=15$ N north. What are the magnitude and direction of the net force $\vec F_{net}$ ?

$35$ N at $37^\circ$ north of east
$25$ N at $53^\circ$ north of east
$25$ N at $37^\circ$ north of east (correct answer)
$5$ N at $37^\circ$ north of east

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 the Pythagorean theorem: magnitude √(20² + 15²) = √(400 + 225) = √625 = 25 N, and direction arctan(15/20) ≈ 37° north of east. Choice C is correct because it properly applies vector addition and correctly determines magnitude and direction. Choice A incorrectly adds the magnitudes directly (20 + 15 = 35), but vectors must be added using 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).

Question 20

In this scenario, a boat's velocity relative to the water is $\vec v_{boat}=\langle 12,0\rangle$ km/h (12 km/h due east). The river current is $\vec v_{current}=\langle 0,5\rangle$ km/h (5 km/h due north). What is the boat's actual velocity relative to the ground (magnitude and direction)?

$13$ km/h due east
$7$ km/h at $\approx 23^\circ$ north of east
$17$ km/h at $\approx 23^\circ$ north of east
$13$ km/h at $\approx 23^\circ$ north of east (correct answer)

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 km/h east and current at 5 km/h north, these are perpendicular, so the resultant velocity magnitude is √(12² + 5²) = √(144 + 25) = √169 = 13 km/h. The direction is arctan(5/12) ≈ 23° north of east. Choice D is correct because it properly applies vector addition and correctly determines magnitude and direction. Choice C incorrectly adds the magnitudes directly (12 + 5 = 17), but vectors must be added using the Pythagorean theorem for perpendicular vectors—you can't just add speeds. Key to relative velocity problems: identify the object's velocity relative to the medium (boat to water) and the medium's velocity relative to the ground, then add these two vectors using components or Pythagorean theorem for perpendicular cases. Physical interpretation: if a boat aims east at 12 km/h but a current flows north at 5 km/h, the boat drifts northeast at 13 km/h, which is what vector addition tells us.