Question

• ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.
During a treasure hunt, Cody walks 45.0 m south and 7.50 m to the west. What single straight-line displacement could
Cody have taken to reach the treasure? (Remember that magnitude and angle and direction are all needed for your
answer)

Answer 1

Answer:

45.6 m at $80.5^{\circ}$ south of west

Explanation:

Let's take the north-south direction as y-direction (with south being positive) and east-west direction as x-direction (with west being positive). Therefore, the two components of Cody's motion are:

- $d_y = 45.0 m$ (south)

- $d_x = 7.50 m$ (west)

Since they are perpendicular, the magnitude of the net displacement can be calculated by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{7.50^2+45.0^2}=45.6 m$

The direction instead can be measured as follows:

$\theta = tan^{-1} (\frac{d_y}{d_x})=tan^{-1}(\frac{45.0}{7.50})=80.5^{\circ}$

And given the convention we have used, this angle is measured as south of west.