Section 1.2: 2D Motion
This section introduces motion in two dimensions, including displacement, velocity, and acceleration vectors. We analyze motion in the x and y directions independently and apply vector addition to solve problems.
Displacement in 2D: \( \vec{r} = x \hat{i} + y \hat{j} \)
Magnitude: \( |\vec{r}| = \sqrt{x^2 + y^2} \)
Direction: \( \theta = \tan^{-1} \frac{y}{x} \)
Velocity components: \( \vec{v} = v_x \hat{i} + v_y \hat{j} \)
Magnitude: \( v = \sqrt{v_x^2 + v_y^2} \)
Direction: \( \theta = \tan^{-1} \frac{v_y}{v_x} \)
Acceleration components: \( \vec{a} = a_x \hat{i} + a_y \hat{j} \)
Magnitude: \( a = \sqrt{a_x^2 + a_y^2} \)
Direction: \( \theta = \tan^{-1} \frac{a_y}{a_x} \)
Example 1
A particle moves with velocity components \(v_x = 3 \text{ m/s}\) and \(v_y = 4 \text{ m/s}\). Find the resultant velocity magnitude and direction.
Magnitude: \( v = \sqrt{3^2 + 4^2} = 5 \text{ m/s} \)
Direction: \( \theta = \tan^{-1} (4/3) \approx 53.13^\circ \) above the x-axis
Practice Problems
- A particle moves 6 m east and 8 m north. Find displacement magnitude and angle.
- A car has velocity components \(v_x = 5 \text{ m/s}, v_y = 12 \text{ m/s}\). Find speed and direction.
- An object moves from (0,0) to (3,4) m. Find displacement vector magnitude and angle.
- Velocity components: \(v_x = -2 \text{ m/s}, v_y = 2 \text{ m/s}\). Find resultant speed and angle.
- A particle accelerates with \(a_x = 1 \text{ m/s²}, a_y = 2 \text{ m/s²}\). Find magnitude and direction of acceleration.