Respuesta :
Answer:
The position is (3.06i-6.34j)= 7.04 Km 64.2 south of east
Explanation:
We can solve this problem by two methods one analytical, using trigonometry and another graph, let's start with the analytical method, we start by decomposing each position vector
V1= 2.50 Km 45.0º norte del oeste
V1x= V1 cos
V1y = Vi sin
All angles must be measured from the positive side of the x axis
45.0º norte del oeste = 180º -45 = 135º
V1x= 2.50 cos 135
V1y = 2.50 sin 135
V1x= -1.77 Km
V1y = 1.77 Km
We repeat this same procedure for all vectors, now we are going to find the angles, let's not forget that all the angles are measured with respect to the positive part of the x-axis counterclockwise
V2= 4.70 Km 60.0º south of east
60.0º south of east = 360- 60 = 300º
V2x= 4.7 cos 300= 2.35 Km
V2y=4.7 sin 300 = -4.07 Km
V3= 1.30 Km 25.0º south of west
25.0º south of west = 180 + 25 = 205º
V3x= 1.3 cos 205= -1.17 Km
V3y= 1.3 sin 205= -0.55 Km
V4= 5.1 Km straight east
East = 0º
V4x= 5.1 Km
V4y=0 Km
V5 = 1.7 Km 5.00º east of north
5.00º east of north = 5º
V5x= 1.7 cos 5= 1.69 Km
V5y= 1.7 sin 5= 0.15 Km
V6= 7.20 Km 55.0º south of west
55.0º south of west = 270-55 = 215º
V6x= 7.2 cos 215=-5.90 Km
V6y= 7.2 sin 215= -4.13 Km
V7= 2.80 Km 10.0º north of east
10.0º north of east = 10º
V7x= 2.80 cos 10= 2.76 Km
V7y= 2.80 sin 10= 0.49 Km
Now we can make the sum for each axis
Vtx = V1x+V2x+V3x+V4x+V5x+V6x+V7x
Vty = V1y+V2y+V3y+V4y+V5y+V6y+V7y
Vtx= -1.77+2.35-1.17+5.1+1.69-5.90+2.76
Vtx = 3.06 Km
Vty= 1.77-4.07-0.55+0+0.15-4..13+0.49
Vty= -6.34 Km
The answer is (3.06i-6.34j), it can also be given as a module and angle
Vt2 = Vtx2 +Vty2
θ = tan-1 (Vty/Vtx)
Vt= 3.062 +(-6.34)2
θ = tan-1 (-6.34/3.06)
Vt= 7.04 Km
θ = - 64.2º
-64.2º = 64.2 south of east
