Let's solve the second problem from the image.

Problem:

• You are kiteboarding eastward at 2.8 m/s when a wind gust hits, lasting 4.2 s, and accelerating your board at 3.6 m/s² at 36° North of East. Assuming you are initially at the origin, determine the magnitude of your final displacement from the origin.

Step-by-step Solution:

1. Initial Velocity:

   • Eastward component: $v_0 = 2.8 \, \text{m/s}$ in the x-direction (East).

2. Wind Gust Acceleration:

   • The wind provides acceleration at 36° North of East, so we resolve the acceleration into its components:
     • $a_x = 3.6 \, \text{m/s}^2 \cdot \cos(36^\circ)$
     • $a_y = 3.6 \, \text{m/s}^2 \cdot \sin(36^\circ)$

3. Final Velocity:

   • To calculate the velocity at the end of 4.2 s:
     • $v_x = v_0 + a_x \cdot t$
     • $v_y = 0 + a_y \cdot t$ (since there was no initial velocity in the y-direction).

4. Displacement:

   • For the x-direction, we use the kinematic equation for displacement: $d_x = v_0 \cdot t + \frac{1}{2} a_x \cdot t^2$
   • For the y-direction: $d_y = \frac{1}{2} a_y \cdot t^2$

5. Magnitude of the Final Displacement:

   • The total displacement is given by: $D = \sqrt{d_x^2 + d_y^2}$

Let me calculate these for you.The magnitude of your final displacement from the origin is approximately 41.84 meters.

Would you like a further breakdown of any step? Here are five related questions you might want to explore:

1. How do we resolve vectors into components?
2. What would happen if the angle of acceleration were different?
3. How do we calculate the velocity after a certain time with constant acceleration?
4. What if the time duration of the wind gust were shorter? How would the displacement change?
5. How does the kinematic equation apply when both initial velocity and acceleration are present?

Tip: In problems involving angles and directions, always break forces or velocities into their components (like x and y) for easier calculation!