Question

1. A particular pass pattern requires a football reciever to run 15m (N) and then 10m (45 degrees W of N). The direction of the resultant displacement is?

Answer 1

Answer on Question #43391 – Physics - Mechanics | Kinematics | Dynamics

A particular pass pattern requires a football receiver to run 15m (N) and then 10m (45 degrees W of N). The direction of the resultant displacement is?

Solution:

We have two displacements: $r_1$ (when receiver moves 15 m North), $r_2$ (when receiver moves 30 m at an angle $45{}^\circ$ West of North) and total displacement $r$.

Displacement along the X-axis:

$$\begin{array}{l} r_{1x} = 0 \, \text{m} \\ r_{2x} = 10 \, \text{m} \cdot \sin(45{}^\circ) = 7 \, \text{m} \\ r_x = r_{1x} + r_{2x} = 0 \, \text{m} + 7 \, \text{m} = 7 \, \text{m} \\ \end{array}$$

Displacement along the Y-axis:

$$\begin{array}{l} r_{1y} = 15 \, \text{m} \\ r_{2y} = 10 \, \text{m} \cdot \sin(45{}^\circ) = 7 \, \text{m} \\ r_y = r_{1y} + r_{2y} = 15 \, \text{m} + 7 \, \text{m} = 22 \, \text{m} \\ \end{array}$$

The direction of the resultant displacement from the right triangle:

$$\tan \theta = \frac{r_x}{r_y} \Rightarrow \theta = \arctan\left(\frac{r_x}{r_y}\right) = \arctan\left(\frac{7 \, \text{m}}{22 \, \text{m}}\right) = 17.7{}^\circ$$

Answer: The direction of the resultant displacement is equal to $17.7{}^\circ$ West of North.

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