Question

I take my dogs on an afternoon walk. We leave my house, walking due East for 4 km. We then walk 3.5 km at an angle of 35 degrees West of North. Finally, we ealk 5.4 km at an angle of 15 degrees West of South. If I were to walk along a straight path that wuld take us back to my house, how far and in what direction should I walk?

Answer 1

In this question we need vector addition of all three displacement, that will give us distance for straight path. Rnet = R1 + R2 + R3

Suppose given that Vector is R and it makes angle A with +x-axis, then it's components are given by:

Rx = R*cos A

Ry = R*sin A

Using above rule:

R1 = 4 km due east (In +ve x-axis)

R1x = 4 km

R1y = 0 km

R2 = 3.5 km at angle 35 deg W of N = 3.5 km at angle 55 deg N of W (2nd Quadrent)

See that we changed angle W of N into N of W by subtracting it from 90 deg

Difference between both of them is that in W of N angle is measured with +y-axis and in N of W angle is mesured with -ve x-axis. Since we defined components w.r.t. angle with x-axis.

R2x = -R2*cos A2 = -3.5*cos 55 deg

R2x = -2.01 km

R2y = R2*sin A2 = 3.5*sin 55 deg

R2y = 2.87 km

R3 = 5.4 km at angle 15 deg W of S = 5.4 km at angle 75 deg S of W

R3x = -R3*cos A3 = -5.4*cos 75 deg

R3x = -1.40 km

R3y = R3*sin A3 = -5.4*sin 75 deg

R3y = -5.22 km

Now

Rnet = Rnet)x + Rnet)y

Rnet = (R1x + R2x + R3x) i + (R1y + R2y + R3y) j

Using above values

Rnet = (4 - 2.01 - 1.40) i + (0 + 2.87 - 5.22) j

Rnet = 0.59 i - 2.35 j

|Rnet| = sqrt (Rnet_x^2 + Rnet_y^2)

|Rnet| = sqrt ((0.59)^2 + (-2.35)^2)

|Rnet| = 2.42 km

Direction of Rnet will be

Direction = arctan (Rnet_x/Rnet_y)

Direction = arctan (2.35/0.59) = 75.91 deg

Direction = 75.91 deg below negative x-axis = 75.91 deg South of East

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