Math Problem Statement

Suppose you walk 11.5 m in a direction exactly 15° south of west then you walk 24 m in a direction exactly 44° west of north.

Solution

To solve this problem, we can break down the two displacements into their respective components and then find the resultant displacement.

Step 1: Breaking the displacements into components

We will use trigonometry to split each displacement into x (horizontal) and y (vertical) components.

Displacement 1: 11.5 m, 15° south of west

  • Since this direction is south of west, the x-component will be negative (west is negative x-direction) and the y-component will also be negative (south is negative y-direction).

    x1=11.5×cos(15)x_1 = 11.5 \times \cos(15^\circ) y1=11.5×sin(15)y_1 = 11.5 \times \sin(15^\circ)

    After calculation, the components are: x1=11.5×cos(15)x_1 = -11.5 \times \cos(15^\circ) y1=11.5×sin(15)y_1 = -11.5 \times \sin(15^\circ)

Displacement 2: 24 m, 44° west of north

  • For this displacement, the x-component will be negative (west) and the y-component will be positive (north).

    x2=24×sin(44)x_2 = 24 \times \sin(44^\circ) y2=24×cos(44)y_2 = 24 \times \cos(44^\circ)

Step 2: Summing the components

Now, sum the x and y components to get the total displacement.

xtotal=x1+x2x_{\text{total}} = x_1 + x_2 ytotal=y1+y2y_{\text{total}} = y_1 + y_2

Step 3: Resultant displacement

To find the magnitude of the resultant displacement, use the Pythagorean theorem:

d=xtotal2+ytotal2d = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}

And to find the direction of the resultant displacement (relative to the positive x-axis):

θ=tan1(ytotalxtotal)\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right)

Let me calculate that.The magnitude of the resultant displacement is approximately 31.24 meters, and the direction is 152.78°, measured counterclockwise from the positive x-axis (or west of north).

Do you have any further questions or would you like more details?

Here are five related questions to expand the topic:

  1. How would the resultant displacement change if the second walk were longer?
  2. How can we use vector addition to solve similar problems?
  3. What is the significance of direction in determining vector components?
  4. How would the displacement differ if we started by walking north of west instead?
  5. How could you find the total time taken if the speed for each walk was given?

Tip: Always remember to consider both magnitude and direction when working with vectors.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Vector Addition
Coordinate Geometry

Formulas

x-component = displacement × cos(angle)
y-component = displacement × sin(angle)
Resultant displacement = √(x_total^2 + y_total^2)
Direction of resultant displacement = tan⁻¹(y_total / x_total)

Theorems

Pythagorean Theorem
Trigonometric Ratios

Suitable Grade Level

Grades 9-12