Unit 1: Advanced Kinematics in Two Dimensions

Reference Frames and Relative Motion

Motion is not absolute; it is defined by your point of view. In physics, this point of view is called a frame of reference.

Concept: It's All Relative

When you state that a car is moving at $20 \text{ m/s}$, you are implicitly making that statement with respect to the ground (the Earth). However, if you were sitting in a different car moving at $20 \text{ m/s}$ right next to it, the first car would appear stationary to you.

This leads to the concept of Relative Velocity.

• Inertial Reference Frame: A frame of reference that is moving at a constant velocity (not accelerating). Newton's laws hold true here without needing fictitious forces.

The Relative Motion Equation

To calculate how fast an object $A$ moves relative to an observer $C$, we use vector addition. The standard notation is $\vec{v}_{AC}$, meaning "the velocity of $A$ with respect to $C$."

The chain rule for relative velocity is:

\vec{v}{AC} = \vec{v}{AB} + \vec{v}_{BC}

Where:

• $\vec{v}_{AC}$ is the velocity of object $A$ relative to frame $C$.
• $\vec{v}_{AB}$ is the velocity of object $A$ relative to frame $B$.
• $\vec{v}_{BC}$ is the velocity of frame $B$ relative to frame $C$.

Note: If you need to reverse a reference frame (e.g., you have $\vec{v}{AB}$ but need $\vec{v}{BA}$), simply flip the sign (direction):
\vec{v}{BA} = -\vec{v}{AB}

Example Scenario: The River Boat

A boat heads North across a river at $4 \text{ m/s}$ relative to the water. The river current flows East at $3 \text{ m/s}$ relative to the ground. What is the velocity of the boat relative to a person standing on the ground?

Solution:

1. Define vectors:

   • $\vec{v}_{b/w} = 4 \text{ m/s}$ (North)
   • $\vec{v}_{w/g} = 3 \text{ m/s}$ (East)
   • Target: $\vec{v}_{b/g}$

2. Apply the equation:
   $\vec{v}{b/g} = \vec{v}{b/w} + \vec{v}_{w/g}$

3. Since these are perpendicular vectors, we define the magnitude using the Pythagorean theorem:
   |\vec{v}_{b/g}| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5 \text{ m/s}

4. Find direction using $\theta = \tan^{-1}(\frac{y}{x})$:
   \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ \text{ North of East}

Vectors and Motion in Two Dimensions

In 1D motion, we used signs ($+$ and $-$) to indicate direction. In 2D, we must analyze motion using components. The "Golden Rule" of 2D Kinematics is:

The horizontal ($x$) motion and vertical ($y$) motion are completely independent of each other.

They share only one variable: Time ($t$).

Vector Decomposition

Any velocity vector $\vec{v}$ launched at an angle $\theta$ relative to the horizontal must be broken down into components.

Projectile Motion

Projectile motion is a form of 2D motion where an object is given an initial velocity and then moves under the influence of gravity alone (ignoring air resistance).

1. Horizontal Characteristics (The X-Axis)

Because gravity acts only vertically ($down$), there is zero acceleration in the horizontal direction.

• $a_x = 0$
• $v_x = \text{constant}$
• Equation: x = x0 + v{ax} t

2. Vertical Characteristics (The Y-Axis)

The object is in free fall.

• $a_y = -g$ (approx $-9.8 \text{ m/s}^2$ or $-10 \text{ m/s}^2$)
• At the peak of the flight, the vertical velocity is momentarily zero ($v_y = 0$).
• Equation: y = y0 + v{oy}t + \frac{1}{2}a_y t^2
• Equation: v{fy} = v{oy} + a_y t

Worked problem: The Cliff Launch

A ball is kicked horizontally off a $20\text{m}$ high cliff with a velocity of $15 \text{ m/s}$. Find the horizontal distance (range) the ball travels.

Solution:

1. Break it down:

   • X-axis: $v{0x} = 15 \text{ m/s}$, $ax = 0$, $x_0 = 0$, find $x$.
   • Y-axis: $y0 = 20\text{m}$, $y = 0$ (ground), $v{0y} = 0 \text{ m/s}$ (launched horizontally), $a_y = -10 \text{ m/s}^2$.

2. Find Time ($t$) using Y-axis data:
   y = y0 + v{oy}t + \frac{1}{2}a_y t^2
   0 = 20 + (0)t + \frac{1}{2}(-10)t^2
   -20 = -5t^2
   t^2 = 4 \quad \Rightarrow \quad t = 2\text{ seconds}

3. Find Range ($x$) using X-axis data and time:
   x = v_{0x} t
   x = (15)(2) = 30\text{ meters}

Common Mistakes & Pitfalls

1. Mixing X and Y variables: This is the most fatal error. You cannot put vertical acceleration ($g$) into a horizontal equation. Never use the total velocity $v$ in a calculation; always use $vx$ or $vy$.

   • Correct: $x = v_x t$
   • Incorrect: $x = v_{total} t$

2. Assuming $v = 0$ at the top: At the peak of a projectile's arc, the vertical velocity $vy$ is zero. However, the horizontal velocity $vx$ is not zero (unless thrown straight up). The object is still moving sideways.

3. Forgetting Negative Signs: Displacement and velocity are vectors. If you define "up" as positive, then gravity ($g$) must be negative ($-9.8 \text{ m/s}^2$), and falling displacement must be negative.

4. Misinterpreting "Dropped" vs. "Thrown":

   • If an object is dropped from a moving plane, its initial horizontal velocity is the same as the plane's velocity ($v{0x} = v{\text{plane}}$).
   • It does not start from rest relative to the ground.

5. Adding Vectors Arithmetically: If velocity A is $3$ and velocity B is $4$, the resultant is only $7$ if they are in the same direction. If they are perpendicular, the resultant is $5$. Always use vectors/Pythagoras.