Set 5 Problem number 9

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

You walk along level ground in a straight line from some initial point to some terminal point.

If I walk .85 miles to the East, then I will be directly South of your terminal point.  Your terminal point will lie 7.19 miles to the North.

• What angle does your path make with East?
• How far did you walk?

Solution

A sketch should consist of two vectors, the first running from the initial point to the East and the other from its terminal point to the North.

• The straight-line path is represented by a vector from the initial point of the first vector to the terminal point of the second.

This vector triangle represents the various displacements:

• The legs of the triangle represent .85 and 7.19 miles, and the hypotenuse represents the actual displacement.
• The straight-line distance is the length of the hypotenuse, which is found by the Pythagorean Theorem to be

• distance = `sqrt( ( .85) ^ 2 + ( 7.19) ^ 2) miles = 7.25 miles.

• The angle at which you should walk will be

• angle = tan^-1 ( 7.19 / .85) = 83.25 degrees.

Generalized Solution

We can set up an x-y coordinate system with the x axis point East and the y axis pointing North.

• If we move through a displacement sx in the x direction then through a displacement sy in the y direction, we will end up at a point that corresponds to a vertex of the triangle formed by applying sx at the starting position, then sy at the terminal point of sx.
• The ending position is at the other end of the hypotenuse from the starting point, and the displacement will be along the hypotenuse.
• The distance is the length of the hypotenuse, which is

• length = distance = `sqrt(sx^2 + sy^2)

• and which lies at angle

• angle = tan^-1(sy / sx)

• from the positive x axis.

Explanation in terms of Figure(s), Extension

The figure below shows the displacements sx and sy as vectors parallel to the x and y axes.

• The path that results from starting at the origin and moving through displacement sx followed by displacement sy is indicated by the heavy (blue) lines forming the legs of the triangle.
• The displacement is the vector `dS from the origin (the starting point of the path) to the terminal point of the path.

The effective distance of this displacement is the length `sqrt(sx^2 + xy^2) of the hypotenuse.

The angle of the displacement is arctan(sy / sx).

Figure(s)