An explorer in the dense jungles of equatorial Africa leaves his hut. He takes 4

Question

An explorer in the dense jungles of equatorial Africa leaves his hut. He takes 42 steps at an angle 45north of east, then 82 steps at an angle 60 north of west, then 53 steps due south. Assume his steps all have equal length. Save him from becoming hopelessly lost in the jungle by giving him the displacement, calculated using the method of components, that will return him to his hut.

What is the magnitude of the displacement that will return the explorer to his hut?

Explanation / Answer

For this problem, you should find the x and y components of each of the steps the explorer took (using SOHCAHTOA), add them up, and then you should end up with the final displacement. We will make north and east positive, so south and west will be negative. 1st step -> 42 steps at 45 degrees D^2 = x1^2 + y1^2, and x1 = y1 since it is a 45 degree angle, 42^2 = 2x1^2, so X1 = 29.70 and Y1 = 29.70 2nd step -> 82 steps 60 degrees north of west (for this, the y component is positive since it is north, but the x component is negative since it is west) sin(60 degrees) = Y2 / 82 Y2 = 71.01 cos(60 degrees) = -X2 / 82 X2 = -41 3rd step -> 53 steps due south (for this, there is no x component, and the y component is south) Y3 = -53 X3 = 0 Total y component = Y1 + Y2 + Y3 = 29.70 + 71.01 - 53 = 47.71 Total x component = X1 + X2 + X3 = 29.70 - 41 + 0 = -11.3 The magnitude of the displacement -> D^2 = X^2 + Y^2 D^2 = 47.71^2 + (-11.3)^2 D = 49.03 steps

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