Question

question 12 (1 point)
which of the following directions is equivalent to n 15° w?
e 75° n
n 75° w
n 75° e
w 75° n
question 13 (1 point)
a student runs 9.0 m w and then turns and runs 12.0 m n. if he is able to do this in 5.0 s, what is his average velocity?
3.0 m/s
4.2 m/s
3.0 m/s w 53° n
4.2 m/s w 53° n
question 14 (1 point)

Explanation:

for Question 12:

Step1: Define [N 15° W]

[N 15° W] means 15° west of north. The angle between north and west is 90°, so the angle from west to north is $90^\circ - 15^\circ = 75^\circ$, which is [W 75° N].

for Question 13:

Step1: Calculate displacement magnitude

Displacement is the straight-line distance from start to end. Using Pythagoras:
$$\text{Displacement} = \sqrt{(9.0\,\text{m})^2 + (12.0\,\text{m})^2} = \sqrt{81 + 144} = \sqrt{225} = 15\,\text{m}$$

Step2: Calculate average velocity magnitude

Average velocity = displacement / time:
$$v_{\text{avg}} = \frac{15\,\text{m}}{5.0\,\text{s}} = 3.0\,\text{m/s}$$

Step3: Find displacement direction

Angle $\theta$ west of north:
$$\tan\theta = \frac{\text{west distance}}{\text{north distance}} = \frac{12.0}{9.0} \approx 1.333 \implies \theta \approx 53^\circ$$
Direction is [W 53° N].

Answer:

Question 12: [W 75° N]
Question 13: 3.0 m/s [W 53° N]