Understanding Motion in 2D

When an object moves in a plane (like the x-y plane), its velocity and acceleration can point in different directions. Unlike 1D motion, here, velocity (\( \vec{v} \)) and acceleration (\( \vec{a} \)) can form any angle between them!

Key Equations for Constant Acceleration

  • Velocity-Time Relation:
    \( \vec{v} = \vec{v}_0 + \vec{a}t \)
    Components: \( v_x = v_{0x} + a_x t \), \( v_y = v_{0y} + a_y t \)
  • Position-Time Relation:
    \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 \)
    Components: \( x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \), \( y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \)

Example Problems

Example 1: Finding velocity and acceleration from position.
Given \( \vec{r} = 3.0t \hat{i} + 2.0t^2 \hat{j} + 5.0 \hat{k} \):
– Velocity: \( \vec{v} = \frac{d\vec{r}}{dt} = 3.0 \hat{i} + 4.0t \hat{j} \)
– Acceleration: \( \vec{a} = \frac{d\vec{v}}{dt} = 4.0 \hat{j} \, \text{m/s}^2 \) (constant in y-direction).

Example 2: A particle starts at the origin with \( \vec{v}_0 = 5.0 \hat{i} \, \text{m/s} \) and acceleration \( \vec{a} = 3.0 \hat{i} + 2.0 \hat{j} \, \text{m/s}^2 \).
– Position at time \( t \): \( x = 5.0t + 1.5t^2 \), \( y = 1.0t^2 \).
– When \( x = 84 \, \text{m} \), solve \( 5.0t + 1.5t^2 = 84 \) → \( t = 6 \, \text{s} \).
– \( y \)-coordinate: \( y = 1.0(6)^2 = 36 \, \text{m} \).
– Speed at \( t = 6 \, \text{s} \): \( \sqrt{(5.0 + 3.0 \times 6)^2 + (2.0 \times 6)^2} = \sqrt{23^2 + 12^2} \approx 26 \, \text{m/s} \).

Projectile Motion

A special case of 2D motion where:
– Horizontal acceleration \( a_x = 0 \) (constant horizontal velocity).
– Vertical acceleration \( a_y = -g \) (due to gravity).
Initial velocity components: \( v_{0x} = v_0 \cos \theta_0 \), \( v_{0y} = v_0 \sin \theta_0 \).

Important Concepts for NEET

  • Independence of Motion: Horizontal and vertical motions are independent. Solve them separately!
  • Projectile Equations: Master \( x = v_{0x}t \), \( y = v_{0y}t – \frac{1}{2}gt^2 \).
  • Vector Magnitude & Direction: Use \( |\vec{v}| = \sqrt{v_x^2 + v_y^2} \) and \( \theta = \tan^{-1}(v_y / v_x) \).
  • Constant Acceleration: Apply \( \vec{v} = \vec{v}_0 + \vec{a}t \) and \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a}t^2 \).

Tip: Practice splitting vectors into components—it simplifies everything! 🚀