Solving Problems with Vectors and Velocity - Precalculus

1. 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 $|\vec v_{ground}|$?
   Answer: $\sqrt{(3)^2+(3)^2}=3\sqrt{2}$ m/s 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).
2. A ball is thrown with initial velocity magnitude $20\ \text{m/s}$ at an angle of $30^\circ$ above the horizontal. For the situation described, what is the horizontal component of the velocity, $v_x$?
   Answer: $20\cos(30^\circ)=10\sqrt{3}\ \text{m/s}$ 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. For the ball thrown at 20 m/s at 30° above horizontal, the horizontal component is $v_x$ = 20 cos(30°) = 20*(√3/2) = 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 for the component calculation, computing 20 sin(30°)=10 m/s instead of 10√3 m/s. 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.
3. For the situation described, a small boat's velocity relative to the water is $\vec v_{\text{boat}}=10\ \text{m/s}$ due west, and the current is $\vec v_{\text{current}}=4\ \text{m/s}$ due east. What is the boat's actual velocity relative to the ground (magnitude and direction)?
   Answer: $6\ \text{m/s}$ due west 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: $\vec{v}_{\text{resultant}} = \vec{v}_{\text{object}} + \vec{v}_{\text{medium}}$. The boat moves at 10 m/s west relative to the water, and the water itself moves at 4 m/s east, so the resultant velocity relative to the ground is the vector sum: $-10 + 4 = -6\ \text{m/s}$ (6 m/s west), since they are collinear and opposite. Choice B is correct because it properly applies vector addition for opposing vectors. Choice A incorrectly adds the magnitudes directly (10 + 4 = 14), but vectors must be added using components or the Pythagorean theorem for perpendicular vectors—you can't just add speeds without direction. 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 west at 10 m/s but a current flows east at 4 m/s, the boat's ground speed is reduced to 6 m/s west, which is what vector addition tells us.
4. Given the velocities, a person walks at $1.5$ m/s due west relative to a train car. The train moves at $10$ m/s due east relative to the ground. How fast is the person moving relative to the ground (magnitude and direction)?
   Answer: $8.5$ m/s due east 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 1.5 m/s west relative to the train, and the train itself moves at 10 m/s east, so to find the resultant velocity relative to the ground, we add these vectors: -1.5 + 10 = 8.5 m/s east. Choice B is correct because it properly interprets relative velocity and correctly adds the vectors along the same line. Choice A incorrectly adds the magnitudes directly (10 + 1.5 = 11.5), but since they are in opposite directions, you must subtract the speeds for collinear vectors. 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, 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.
5. 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)?
   Answer: $13\ \text{m/s}$ at $\tan^{-1}(5/12)\approx 22.6^\circ$ east of north 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.
6. 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)?
   Answer: $13\ \text{km/h}$ at $\tan^{-1}(5/12)$ east of north 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.
7. In this scenario, two forces act on a crate: $\vec F_1=20\text{ N}$ east and $\vec F_2=15\text{ N}$ north. What are the magnitude and direction of the net force $\vec F_{\text{net}}$ (direction as an angle north of east)?
   Answer: $25\text{ N}$ at $\tan^{-1}\!\left(\frac{15}{20}\right)$ north of east 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 crate. Since they are perpendicular, we use the Pythagorean theorem: |F⃗_net| = √(20² + 15²) = √(400 + 225) = √625 = 25 N. The direction is arctan(15/20) north of east. Choice C is correct because it properly applies vector addition using the Pythagorean theorem for perpendicular forces and correctly determines both magnitude (25 N) and direction (arctan(15/20) north of east). Choice B gives the correct magnitude but uses the wrong ratio for the angle, computing arctan(20/15) instead of arctan(15/20), which would give the angle east of north rather than north of east. 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.
8. 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?
   Answer: $5\text{ m/s}$ 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.
9. For the situation described, a ball is thrown with initial velocity magnitude $20\ \text{m/s}$ at an angle of $30^\circ$ above the horizontal. What is the horizontal component of the velocity, $v_x$?
   Answer: $20\cos(30^\circ)=10\sqrt{3}\ \text{m/s}$ 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).
10. 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)?
    Answer: $\tan^{-1}(30/40)\approx 36.9^\circ$ east of north 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.
11. Given the velocities, a hiker walks with velocity $\vec v_1=\langle 3,4\rangle\ \text{m/s}$ (components in east, north), and a moving walkway adds $\vec v_2=\langle 2,0\rangle\ \text{m/s}$. What is the velocity of the hiker relative to the ground, $\vec v_{\text{ground}}$?
    Answer: $\langle 5,4\rangle\ \text{m/s}$ 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: velocity 1 has components ⟨3, 4⟩ and velocity 2 has components ⟨2, 0⟩. Adding: v⃗_result = ⟨3 + 2, 4 + 0⟩ = ⟨5, 4⟩. Choice B is correct because it properly applies vector addition using components. Choice A makes an arithmetic error, calculating ⟨3-2, 4⟩ = ⟨1,4⟩ instead of ⟨3+2,4⟩. 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). Key to relative velocity problems: identify the object's velocity relative to the medium and the medium's velocity relative to the ground, then add these two vectors using components or Pythagorean theorem for perpendicular cases.
12. 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?
    Answer: $0\ \text{m/s}$ (the zero vector) 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.
13. In this scenario, two forces act on a crate: $\vec F_1=20\ \text{N}$ east and $\vec F_2=15\ \text{N}$ north. What are the magnitude and direction of the net force $\vec F_{\text{net}}$?
    Answer: $25\ \text{N}$ at $\tan^{-1}(15/20)\approx 37^\circ$ north of east 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).
14. 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, $v_x$?
    Answer: $20\cos(30^\circ)\text{ m/s}$ 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.
15. Given the velocities, a person walks at $\vec v_{\text{person/train}}=1.5\ \text{m/s}$ east relative to a train, and the train moves at $\vec v_{\text{train/ground}}=10\ \text{m/s}$ east relative to the ground. How fast is the person moving relative to the ground?
    Answer: $11.5\ \text{m/s}$ east 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: $\vec{v}_{\text{resultant}} = \vec{v}_{\text{object}} + \vec{v}_{\text{medium}}$. The person moves at 1.5 m/s east relative to the train, and the train itself moves at 10 m/s east, so the resultant velocity relative to the ground is the vector sum: $10 + 1.5 = 11.5$ m/s east, since they are parallel. Choice C is correct because it properly applies vector addition for collinear vectors. Choice D treats the velocities as perpendicular when they are parallel, incorrectly using $\sqrt{10^2 + 1.5^2}$ instead of simple addition. 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, then add these two vectors using components or Pythagorean theorem for perpendicular cases. Physical interpretation: if a person walks east on an eastward-moving train, their ground speed is faster than either alone, which is what vector addition tells us.
16. 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)?
    Answer: $37^\circ$ north of west 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.
17. 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 $\vec F_{net}$?
    Answer: $25$ N at $37^\circ$ north of east 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).
18. For the situation described, a ball is thrown with initial velocity magnitude $20$ m/s at $30^\circ$ above the horizontal. What is the horizontal component of the velocity (ignore air resistance)?
    Answer: $20\cos(30^\circ)=10\sqrt{3}$ m/s 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.
19. 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?
    Answer: 2.0 km/h 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.
20. Two forces act on a crate on a frictionless floor: $\vec F_1=20\ \text{N}$ east and $\vec F_2=15\ \text{N}$ north. Given the forces, what are the magnitude and direction of the net force $\vec F_{\text{net}}$?
    Answer: $25\ \text{N}$ at $37^\circ$ north of east 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 B 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 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). 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.

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