solve each with solution each ACTIVITY 1FINDING RESULTANT VECTOR USING GRAPHICAL METHOD- POLYGON METHOD1. Displacementd1= 7 m Sd2= 6 m 40 degree NEd3= 7 m Ed4= 6 m 55 degree NWFind dR, angle and directionScale: 1 m=1 cm2. Velocityv1= 70 km/hr Wv2= 50 km/hr 29 degree SWv3= 80 km/hr 66 degree SEv4= 90 km/hr 50 degree NEFind vR, angle and directionScale: 10 km/hr= 1 cm3. ForceF1= 300 N 80 degree NEF2= 500 N 25 degree NWF3= 300 N 40 degree SWF4= 400 N EastFind: FR, angle and directionScale: 100 N= 1 inch​

Answer:ACTIVITY 1 — DisplacementScale: 1 m = 1 cmCoordinate convention: +x = East, +y = North.Vectors (components):d₁ = 7 m S → 1=0, 1=−7d1x ​=0, d1y ​=−7d₂ = 6 m, 40° NE (40° east of north)2=6sin⁡40∘=3.8567 m,  2=6cos⁡40∘=4.5963 md2x ​=6sin40∘=3.8567 m,d2y ​=6cos40∘=4.5963 md₃ = 7 m E → 3=7, 3=0d3x ​=7, d3y ​=0d₄ = 6 m, 55° NW (55° west of north)4=−6sin⁡55∘=−4.9149 m,  4=6cos⁡55∘=3.4415 md4x ​=−6sin55∘=−4.9149 m,d4y ​=6cos55∘=3.4415 mSum components:=0+3.8567+7−4.9149=5.9418 mdRx ​ ​=0+3.8567+7−4.9149=5.9418 m=−7+4.5963+0+3.4415=1.0377 mdRy ​ ​=−7+4.5963+0+3.4415=1.0377 mMagnitude and direction:∣∣=5.94182+1.03772=6.03 m∣dR ​∣=5.94182+1.03772 ​=6.03 m (rounded)Angle from +x (east): =arctan⁡ ⁣(1.03775.9418)=9.91∘θ=arctan(5.94181.0377 ​)=9.91∘ → 9.9° N of EAnswer (Displacement):=6.03 m,  9.9∘ north of eastdR ​=6.03 m,9.9∘ north of east ​Graphical (scale): draw a line of length 6.03 cm (since 1 m = 1 cm) at 9.9° above the east direction.ACTIVITY 2 — VelocityScale: 10 km/h = 1 cmWe keep same axes (+x east, +y north).Vectors:v₁ = 70 km/h W → 1=−70, 1=0v1x ​=−70, v1y ​=0v₂ = 50 km/h, 29° SW (29° west of south)2=−50sin⁡29∘=−24.2405,  2=−50cos⁡29∘=−43.7310v2x ​=−50sin29∘=−24.2405,v2y ​=−50cos29∘=−43.7310v₃ = 80 km/h, 66° SE (66° east of south)3=80sin⁡66∘=73.0836,  3=−80cos⁡66∘=−32.5389v3x ​=80sin66∘=73.0836,v3y ​=−80cos66∘=−32.5389v₄ = 90 km/h, 50° NE (50° east of north)4=90sin⁡50∘=68.9440,  4=90cos⁡50∘=57.8509v4x ​=90sin50∘=68.9440,v4y ​=90cos50∘=57.8509Sum components:=−70−24.2405+73.0836+68.9440=47.7872 km/hvRx ​ ​=−70−24.2405+73.0836+68.9440=47.7872 km/h=0−43.7310−32.5389+57.8509=−18.4190 km/hvRy ​ ​=0−43.7310−32.5389+57.8509=−18.4190 km/hMagnitude & direction:∣∣=47.78722+(−18.4190)2=51.21 km/h∣vR ​∣=47.78722+(−18.4190)2 ​=51.21 km/hAngle from +x: =arctan⁡ ⁣(−18.419047.7872)=−21.08∘θ=arctan(47.7872−18.4190 ​)=−21.08∘ → 21.1° S of EAnswer (Velocity):=51.21 km/h,  21.1∘ south of eastvR ​=51.21 km/h,21.1∘ south of east ​Graphical (scale): at 10 km/h = 1 cm, draw resultant length = 51.21/10=5.121 cm51.21/10=5.121 cm at 21.1° below east.ACTIVITY 3 — ForceScale: 100 N = 1 inchAxes same (+x east, +y north).Vectors:F₁ = 300 N, 80° NE (80° east of north)1=300sin⁡80∘=295.4423,  1=300cos⁡80∘=52.0945F1x ​=300sin80∘=295.4423,F1y ​=300cos80∘=52.0945F₂ = 500 N, 25° NW (25° west of north)2=−500sin⁡25∘=−211.3091,  2=500cos⁡25∘=453.1539F2x ​=−500sin25∘=−211.3091,F2y ​=500cos25∘=453.1539F₃ = 300 N, 40° SW (40° west of south)3=−300sin⁡40∘=−192.8363,  3=−300cos⁡40∘=−229.8133F3x ​=−300sin40∘=−192.8363,F3y ​=−300cos40∘=−229.8133F₄ = 400 N East → 4=400, 4=0F4x ​=400, F4y ​=0Sum components:=295.4423−211.3091−192.8363+400=291.2969 NFRx ​ ​=295.4423−211.3091−192.8363+400=291.2969 N=52.0945+453.1539−229.8133+0=275.4350 NFRy ​ ​=52.0945+453.1539−229.8133+0=275.4350 NMagnitude & direction:∣∣=291.29692+275.43502=400.90 N∣FR ​∣=291.29692+275.43502 ​=400.90 NAngle from +x: =arctan⁡ ⁣(275.4350291.2969)=43.40∘θ=arctan(291.2969275.4350 ​)=43.40∘ → 43.4° N of EAnswer (Force):=400.9 N,  43.4∘ north of eastFR ​=400.9 N,43.4∘ north of east ​Graphical (scale): at 100 N = 1 in, draw resultant length = 400.9/100=4.009 in400.9/100=4.009 in (≈ 4.01 inches) at 43.4° above east.How to draw each result using the polygon (graphical) methodPick your paper axes (East to the right, North upward). Mark the scale given for that question.Draw each vector head-to-tail in sequence (use a protractor for the angle and ruler for length at scale). Example: for Activity 1, draw d₁ (7 cm south), then from d₁’s head draw d₂ (6 cm at 40° NE), etc.After placing the last vector, draw a straight line from the tail of the first vector to the head of the last vector — that line is the resultant.Measure its length (convert back using the scale) and measure angle from the +x axis (east) to get the direction (state as N of E, S of E, etc).

Answered by canetedaniel | 2025-08-26