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Unit 1, Assignment 1- Carrie Richards

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School
Durham College**We aren't endorsed by this school
Course
PHYSICS 101
Subject
Physics
Date
Jan 2, 2025
Pages
6
Uploaded by DukeClover8405
Unit 1, Assignment 1PhysicsCarrie Richards1. A person drives 8.0 km [N], 6.0 km [W]. Find the total displacement using thePythagorean theorem and trigonometry. Include a diagram. (5 marks)Formula: d= √(a²+b²)a= 8.0 km, b=6.0 kmd=√(8.0)² + (6.0)² =√64+36 = √100 = 10.0 kmtan(θ) =adjacent/opposite = 6.0/8.0θ=tan𐄐¹(6.0/8.0) ≈ 36.87॰The direction is [N 36.87॰W]The total displacement is 10.0 km [N36.87॰W]Diagram(s):2. A person travels 2.0 m [E 20° S], then 4.0 m [S]. Find the total displacement usingtrigonometry (sine and cosine laws). Include a diagram.East component = 2.0 m cos(20°) ≈ 1.8794 m
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South component = 2.0 m sin(20°) ≈ 0.6840 mTotal South component = 0.6840 m + 4.0m = 4.6840 mThe east component stays the same:East component = 1.8794 kmUsing the Pythagorean theorem:d = √(1.8794 m)² + (4.6840 m)²d = √3.5321 + 21.9378 ≈ 5.0467 mtanθ = 4.6840m/1.8784θ = tan𐄐¹(4.6840/1.8794) ≈ 68.20°Total displacement is 5.0467 m [E68.20°S]Diagram:||(N)|↑|||| 20 km||||__________(W 30° S)|/|/|/ 80 km|/|/|/|/| /| /|/----------------------→ (S)100 km
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3.A truck drives 1.0 x 102 km [S], turns and drives 80.0 km [W 30° S], then turns againand drives 20.0 km [N]. Find the total displacement using the perpendicular componentsmethod. Include a diagram.100 km [S] = 0 km E, 100 km SFor 80 km [W 30°S]:E = 80 cos(30°) ≈ 69.28 km WS = 80 sin(30°≈ 40 km SFor 20 km [N]:E = 0 , S = 20 km NEast-West: - 69.28 kmNorth-South: 100 + 40 - 20 = 120 km Sd = √(69.28² + 120²) ≈ √(4800 + 14400) ≈ √19200 ≈ 138.59 kmθ = tan𐄐¹(120/69.28) ≈ 59.04° S of WThe total displacement is 138.59 km at 59.04° S of WDiagram:4. A canoeist aims their canoe [N30◦W] and paddles at a speed of 3.0 m/s to cross a riverthat is flowing East at 0.80 m/s. What is the velocity of the canoe relative to theriverbank? Include a diagram.Canoe’s direction: N30°WCanoe’s speed:3.0 m/sRiver's speed: 0.80 m/s (E)
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Canoe velocity components:North component (Vcy):Vcy= 3.0 X sin(30°) = 3.0 X 0.5 = 1.5 m/s (N)West component (Vcx):Vcx= -3.0 X cos(30°) = -3.0 X √3/2 ≈ -2.6 m/s (W)River Velocity:East Component (Vrx):Vrx= 0.80 m/s (E)North component (Vry):Vry=0 m/sTotal X component (East-West):Vtotalx =Vcx +Vrx= -2.6 + 0.80 = -1.8 m/s (W)Vtotaly= Vcy + Vry= 1.5 + 0 = 1.5 m/s (N)Using the Pythagorean Theorem:Vtotal= √(Vtotalx)² + Vtotaly)²Vtotal= √(-1.8)² + (1.5)² = √3.24 + 2.25 = √5.49 ≈ 2.34 m/sθ = tan𐄐¹ (Vtotaly/Vtotalx= tan𐄐¹ (1.5/1.8) ≈ 39.8° N of WThe velocity of the canoe relative to the riverbank is approximately 2.34 m/s at an angle of 39.8°N of W.Diagram:5. An airplane flies with an airspeed of 50.0 m/s [E 40° N]. If the velocity of the airplane,according to an observer on the ground, is 30.0 m/s [SE], what is the wind velocity?Include a diagram
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Airspeed of airplane: Va= 50.0 m/s [E 40° N]Ground speed of the airplane: Vg= 3.0 m/s [SE]Airplane velocity components:East component: (Vax):Vax= 50.0 X cos(40°) ≈ 50.0 X 0.766 = 38.3 m/sNorth component (Vay):Vay = 50.0 X sin(40°)≈ 50.0 X 0.643 = 32.2 m/sGround velocity components:East component (Vgx):Vgx =3.0 X cos(45°) ≈ 30.0 X 0.707 = 21.2 m/sSouth component (Vgy):Vgy= -30.0 X sin(45°) ≈ -30.0 X 0.707 = -21.2 m/sUsing the equation: Vg= Va +VwWhere Vwis the wind velocity vector. Rearranging gives: Vw= Vg- VaNorth component (Vwy):Vwy= Vgy- Vay= -21.2 - 32.2 = -53.4 m/sMagnitude:Vw= √(Vw)² + (Vwy)² = √(-17.1)² + (-53.4)² ≈ √292.41 + 2855.56 ≈ √3147.97 ≈ 56.1 m/sDirection:θ = tan𐄐¹(Vwy/Vwx) = tan𐄐¹(53.4/17.1) ≈ 72.3° S of WThe wind velocity is approximately 56.1 m/s at an angle of 72.3° S of WDiagram:N||\| \| \ Airplane Velocity (50 m/s [E 40° N])|\|\|\|\|\|---------> Ground Speed (30 m/s [SE])||+------------------> E
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