Applied Vector Word Problems

What you’ll use

Components Break any 2D vector into x (east–west) and y (north–south).
Addition Add components: ⟨a_x,a_y⟩+⟨b_x,b_y⟩=⟨a_x+b_x,a_y+b_y⟩.
Magnitude |v| = √(v_x²+v_y²).
Angles (SOH–CAH–TOA) with θ from +x: v_x=|v|cosθ, v_y=|v|sinθ, and θ = atan2(v_y, v_x).
Average speed/velocity speed = distance / time, velocity = displacement / time (vector).

Tip: State your coordinate system and draw a quick diagram before you compute.

Practice Set (20 problems)

Work each problem. Open the solution for a check and a short walkthrough.

Airplane and wind vectors. An airplane flies at 500 km/h due north through a 50 km/h wind blowing from the west. (a) What is the plane’s ground-speed and track? (b) If the pilot wants to fly exactly north, what heading is required and what ground speed results?
Show solution
1. Choose axes: +x = east, +y = north. Airplane (airspeed) ⟨0, 500⟩ km/h. Wind from west ⇒ ⟨+50, 0⟩ km/h.
2. Ground velocity: ⟨50, 500⟩ km/h.
3. Magnitude: |v| = √(50² + 500²) ≈ 502.5 km/h.
4. Direction (east of north): tanα = 50/500 ⇒ α ≈ 5.71°.
5. (b) To track due north, cancel wind’s east component: need airplane’s **west** component 50 km/h. Let δ be the heading angle west of north.
6. Set 500·sinδ = 50 ⇒ δ = arcsin(0.1) ≈ 5.74° (west of north).
7. Ground speed is then the north component: 500·cosδ ≈ 497.5 km/h due north.
Answer. (a) 502.5 km/h, 5.71° east of north. (b) Heading 5.74° west of north; ground speed 497.5 km/h north.
River current and boat. River flows east at 3 m/s. Boat speed is 4 m/s relative to water. To land directly north, what heading and what bank-relative speed?
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1. Let the boat aim φ west of north so its **west** component cancels the current.
2. Component match: 4·sinφ = 3 ⇒ φ = arcsin(3/4) ≈ 48.59° (west of north).
3. Northward crossing speed: v = 4·cosφ ≈ 2.65 m/s (since east–west cancels).
Answer. Heading 48.59° west of north; bank-relative speed 2.65 m/s north.
Hiker’s journey. 2 km east, 4 km south, 5 km west.
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1. Total distance: 2 + 4 + 5 = 11 km.
2. Displacement: ⟨2−5, 0−4⟩ = ⟨−3, −4⟩ km.
3. Magnitude: |Δr| = √(3² + 4²) = 5.00 km.
4. Direction: tanθ = 4/3 ⇒ θ ≈ 53.13° south of west.
Answer. Distance 11 km. Displacement 5.00 km, 53.13° south of west.
Robot vacuum. 5 m east, 3 m north, 2 m west.
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1. Displacement: ⟨5−2, 3⟩ = ⟨3, 3⟩ m.
2. Magnitude: √(3²+3²) = √18 ≈ 4.24 m.
3. Direction: components equal ⇒ 45° north of east.
Answer. 4.24 m at 45° north of east.
Car’s average speed vs. velocity. 10 km east in 15 min, then 2 km west in 5 min.
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1. Times: 15 min = 0.25 h; 5 min = 0.0833 h; total time = 0.3333 h.
2. Total distance = 12 km ⇒ average speed = 12 / 0.3333 ≈ 36 km/h.
3. Displacement = 10 − 2 = 8 km east ⇒ average velocity = 8 / 0.3333 ≈ 24 km/h east.
Answer. Avg speed 36 km/h. Avg velocity 24 km/h east.
Submarine navigation. Sub: 20 kt on bearing 120°. Current: 5 kt on bearing 225°. Find resultant (magnitude & bearing).
Show solution
1. Convert bearing β (clockwise from north) to components: ⟨v_E, v_N⟩ = ⟨v·sinβ, v·cosβ⟩.
2. Sub (120°): ⟨20·sin120°, 20·cos120°⟩ = ⟨17.32, −10.00⟩ kt.
3. Current (225°): ⟨5·sin225°, 5·cos225°⟩ = ⟨−3.54, −3.54⟩ kt.
4. Sum: ⟨13.79, −13.54⟩ kt.
5. Magnitude: √(13.79² + 13.54²) ≈ 19.32 kt.
6. Bearing (clockwise from north): θ = atan2(E, N) ≈ atan2(13.79, −13.54) ≈ 134.5°.
Answer. 19.32 kt on bearing 134.5°.
Swimmer (one length). One 50 m length in 25 s.
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1. Average speed = distance / time = 50 / 25 = 2.00 m/s.
2. Average velocity magnitude for one length = 2.00 m/s toward the far wall (direction constant).
3. Down-and-back in the same time would give zero displacement ⇒ average velocity 0, but average **speed** would remain distance/time.
Answer. 2.00 m/s toward the far wall.
Delivery driver. 3 mi north, 4 mi east in 15 min. Find average speed and average velocity.
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1. Total distance: 3 + 4 = 7 mi; time: 0.25 h.
2. Average speed: 7 / 0.25 = 28 mph.
3. Displacement: ⟨4, 3⟩ mi ⇒ magnitude √(4²+3²) = 5 mi.
4. Average velocity magnitude: 5 / 0.25 = 20 mph, at 53.13° east of north.
Answer. Avg speed 28 mph; avg velocity 20 mph at 53.13° east of north.
River crossing (constant velocity). A river is 300 m wide (banks run east–west). The current is 2.0 m/s east. A boat can move at 4.0 m/s relative to the water and aims due north.
Show solution
1. Choose axes: +x = east, +y = north. Boat rel. water: ⟨0, 4.0⟩ m/s. Current: ⟨2.0, 0⟩ m/s.
2. Ground velocity: ⟨2.0, 4.0⟩ m/s.
3. Time to cross (width / north component): t = 300 / 4.0 = 75 s.
4. Downstream drift: x = v_x t = 2.0 × 75 = 150 m east.
5. Track angle relative to north: tanα = 2.0/4.0 = 0.5 ⇒ α ≈ 26.6° east of north.
Answer. Time 75 s, drift 150 m east, track 26.6° east of north.
Tugboat forces. A: 5000 N @ 30° N of E. B: 7000 N @ 60° S of E. Resultant?
Show solution
1. A components: ⟨5000 cos30°, 5000 sin30°⟩ = ⟨4330.13, 2500.00⟩ N.
2. B components: ⟨7000 cos60°, −7000 sin60°⟩ = ⟨3500.00, −6062.18⟩ N.
3. Sum: ⟨7830.13, −3562.18⟩ N.
4. Magnitude: |F| = √(7830.13² + 3562.18²) ≈ 8602.3 N.
5. Direction: θ = atan(|−3562.18| / 7830.13) ≈ 24.5° south of east.
Answer. 8602.3 N at 24.5° south of east.
Two walkers, perpendicular paths (relative velocity). Alex walks north at 1.5 m/s along 1st Ave; Jamie walks east at 1.0 m/s along Pine St. They start 420 m south and 560 m west of the same intersection at the same time. Will they reach the intersection together? If not, who arrives first and by how much?
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1. Times to intersection (constant speeds): t_A = 420/1.5 = 280 s; t_J = 560/1.0 = 560 s.
2. They do not arrive together.
3. Alex arrives first. Time gap: Δt = 560 − 280 = 280 s (4 min 40 s).
4. Optional: relative velocity magnitude is √(1.5²+1.0²) ≈ 1.80 m/s, but arrival is set by individual times here.
Answer. Not together; Alex first by 280 s (4:40).
Wind triangle (constant airspeed). A drone’s airspeed is 12.0 m/s. The wind blows 4.0 m/s toward the east. The pilot wants a ground track due north. What heading (west of north) and what ground speed?
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1. Let heading be δ west of north so air’s west component cancels wind: 12·sinδ = 4 ⇒ sinδ = 1/3.
2. Heading: δ = arcsin(1/3) ≈ 19.47° west of north.
3. Ground speed is north component: v = 12·cosδ = 12·√(1−(1/3)²) = 12·(2√2/3) ≈ 11.31 m/s.
Answer. Heading 19.47° west of north; ground speed ≈ 11.31 m/s north.
Wind-assisted drone. 10 m/s north with crosswind 3 m/s east.
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Ground speed = \(\sqrt{10^2+3^2}\) ≈ 10.44 m/s; direction ≈ 16.7° east of north.
Mountain climber. Base → A: 1000 m east, +500 m up. A → B: 800 m east, +300 m up.
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Base → B: ⟨1800, 800⟩ m. |Δr| ≈ 1969.8 m.
Soccer cut. 15 m north, then 10 m west.
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|Δr| = \(\sqrt{15^2+10^2}\) ≈ 18.03 m at 33.7° west of north.
Carrier take-off. Carrier 20 kt south; aircraft 150 kt south relative to ship.
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Relative to water: 170 kt south.
Relative velocity at intersection. Car: 30 mph north. Truck: 40 mph east. Velocity of car relative to truck?
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\(\vec v_{c/t} = \vec v_c - \vec v_t = ⟨-40, 30⟩\) mph ⇒ magnitude 50 mph at 36.9° north of west.
Dog sled bearings. 5 km @ 45° (NE), then 3 km @ 135° (SE).
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Δr ≈ ⟨5.66, 1.41⟩ km ⇒ |Δr| ≈ 5.83 km, bearing ≈ 14.0° north of east.
Passenger on a train. Train 60 m/s east; passenger runs 5 m/s west relative to train.
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Ground frame: 55 m/s east.
Piecewise constant trip (average speed vs. velocity). A courier drives 6 km east in 12 min, waits 3 min, then drives 2 km north in 5 min. Find the average speed and the average velocity for the whole trip.
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1. Total time: 12 + 3 + 5 = 20 min = 0.333 h.
2. Total distance: 6 + 0 + 2 = 8 km ⇒ average speed 8 / 0.333 ≈ 24.0 km/h.
3. Displacement: ⟨6, 2⟩ km ⇒ magnitude √(6²+2²) = √40 ≈ 6.3249 km.
4. Average velocity magnitude: 6.3249 / 0.333 ≈ 19.0 km/h, direction tanθ = 2/6 ⇒ θ ≈ 18.4° north of east.
Answer. Avg speed ≈ 24.0 km/h. Avg velocity ≈ 19.0 km/h at 18.4° north of east.

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