Section 1.2: Vectors and 2D Motion

In two-dimensional motion, objects move in a plane. To describe such motion, we use vectors, which have both magnitude and direction, unlike scalars which only have magnitude.

        Scalars vs Vectors:
        Scalars: Only magnitude (speed, distance, mass, temperature).
Vectors: Magnitude and direction (velocity, displacement, acceleration, force).

      

        Vector Representation:
        Arrow pointing in direction of the vector.
Length proportional to magnitude.
Components along x and y axes: \( \vec{A} = A_x \hat{i} + A_y \hat{j} \)

      

Vector Addition and Subtraction

Vectors can be combined using either the graphical method or components method.

        Graphical (Tip-to-Tail) Method: Place the tail of the second vector at the tip of the first vector. The resultant vector goes from the tail of the first to the tip of the last.
      

        Component Method: Break each vector into x and y components and add algebraically:
        \[
        R_x = A_x + B_x, \quad R_y = A_y + B_y
        \]
        \[
        |\vec{R}| = \sqrt{R_x^2 + R_y^2}, \quad \theta_R = \tan^{-1}\frac{R_y}{R_x}
        \]
      

Resolving Vectors into Components

Given a vector magnitude \(A\) at an angle \(\theta\) from the x-axis:

\(A_x = A \cos \theta\)
\(A_y = A \sin \theta\)

This allows us to handle motion along perpendicular directions independently.

2D Motion Concepts

For motion in a plane with constant velocity in x and y:

\(x = x_0 + v_x t\)
\(y = y_0 + v_y t\)

This is general 2D motion without gravity or other forces.

Example 1

A boat moves 3 km east and then 4 km north. Find the resultant displacement and its direction.

Magnitude: \( |\vec{R}| = \sqrt{3^2 + 4^2} = 5 \text{ km} \)

Direction: \( \theta = \tan^{-1}\frac{4}{3} \approx 53.1^\circ \) north of east

Example 2

A particle moves with velocity components \(v_x = 6 \text{ m/s}\), \(v_y = 8 \text{ m/s}\). Find the magnitude and direction of its velocity.

\(v = \sqrt{6^2 + 8^2} = 10 \text{ m/s}\)

\(\theta = \tan^{-1}(8/6) \approx 53.1^\circ\) above x-axis

Practice Problems

A displacement vector of 5 m east and 12 m north. Find resultant displacement and angle.
Two vectors: \(A = 10 \hat{i} + 5 \hat{j}\), \(B = -3 \hat{i} + 7 \hat{j}\). Find \(A+B\) and its magnitude.
A particle moves 8 m at 30° to x-axis, then 6 m at 120° to x-axis. Find total displacement.
Given \(v_x = 3\) m/s, \(v_y = -4\) m/s, find speed and direction.
Resolve a 10 N force at 60° above horizontal into x and y components.
A hiker walks 4 km north, then 3 km east. Determine resultant displacement.
Particle moves in x-y plane: \(x = 2t\), \(y = 3t\). Find magnitude and direction of velocity at t=2 s.
Two vectors at right angles: 6 N east and 8 N north. Find magnitude and direction of resultant.
A drone moves 50 m at 45°, then 70 m at 135°. Find resultant displacement.
Vector \(A = 7 \hat{i} - 24 \hat{j}\), find magnitude and angle with x-axis.

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