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Question

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An object was subjected to two displacement vectors. Vector One,V1, had a magnitude of 426 m and was directed at an angle of 40 degrees North of East, while Vector Two,V2, had a magnitude of 125 m and was directed at an angle of 68 degrees North of East. Calculate the resultant of the addition of these two vectors.

Give your answer in metres and degrees and round both values to the nearest whole value using the normal convention.

Resultant displacement (m) =

Angle, North of East (degrees) =

Answer 1

540 m

46 degrees North of East

Answer 2

Separate both vectors into x and y components.

Sum up each component.

Find the resultant vector.

In the following, 0° is taken to be East

V1x + V2x =

426 cos 40° + 125 cos 68° =

326.33 + 46.83 = 373.16

V1y + V2y =

426 sin 40° + 125 sin 68° =

273.83 + 115.90 = 389.72

magnitude of net displacement =

sqrt ( 373.16 ^2 + 389.72 ^2 ) = 540 m

angle =

arctan (389.72 / 373.16) = 46° north of east

Answer 3

V1 + V2 have each  2components,

V1x=426*cos(40), V1y=426*sin(40)

V2x=125*cos(68), V2y=125*sin(68)

Vx = V1x + V2x  =326.3 + 46.8 = 373.2

Vy = V1y + V2y  =273.8 +115.9 = 389.7

  

Vr = sqrt(Vx^2+Vy^2) =539.6 m

Angle = tan^-1 O = tan^-1 (389.7/373.2) = tan^-1(1.04) =46.2 Degr from X toward Y axes counterclockwise or North of East