Position and Displacement

Position

Position ($x$) is the location of an object relative to a reference point (origin).

• Measured in meters (m)
• Can be positive or negative
• Example: $x = 5$ m means 5 meters to the right of the origin

Displacement

Displacement ($\Delta x$) is the change in position.

$\Delta x = x_f - x_i$

Where:

• $x_f$ = final position
• $x_i$ = initial position

Note: Displacement is different from distance traveled!

• Displacement: straight-line change (can be negative)
• Distance: total path length (always positive)

Velocity

Average Velocity

Average velocity is displacement divided by time interval.

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$

• Units: m/s
• Can be positive (moving right/up) or negative (moving left/down)
• Vector quantity (has direction)

Instantaneous Velocity

Instantaneous velocity is the velocity at a specific instant.

$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

• This is the slope of the position-time graph at a point
• On a graph: tangent line slope

Speed

Speed is the magnitude of velocity (always positive).

$\text{speed} = |v|$

Acceleration

Average Acceleration

Average acceleration is the change in velocity divided by time.

$a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$

• Units: m/s²
• Can be positive or negative
• Positive $a$: speeding up in positive direction OR slowing down in negative direction

Instantaneous Acceleration

Instantaneous acceleration is the acceleration at a specific instant.

$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

• This is the slope of the velocity-time graph at a point

Kinematic Equations (Constant Acceleration)

When acceleration is constant, we can use these powerful equations:

The Big Four Kinematic Equations

$v = v_0 + at$

$x = x_0 + v_0 t + \frac{1}{2}at^2$

$v^2 = v_0^2 + 2a(x - x_0)$

$x = x_0 + \frac{1}{2}(v_0 + v)t$

Where:

• $x_0$ = initial position
• $x$ = final position
• $v_0$ = initial velocity
• $v$ = final velocity
• $a$ = acceleration (constant)
• $t$ = time

Choosing the Right Equation

Free Fall

Free fall is motion under gravity alone (no air resistance).

$a = -g = -9.8 \text{ m/s}^2$

• Negative because gravity pulls downward
• Same equations apply, but use $a = -g$
• Objects thrown upward: velocity decreases at rate $g$ until $v = 0$ at peak

Key Facts About Free Fall

1. At the peak of motion, $v = 0$ but $a = -g$
2. Time up = time down (for same height)
3. Landing speed = launching speed (magnitude)

Sign Conventions

Always establish a coordinate system first!

Common convention:

• Positive direction: right, up
• Negative direction: left, down

Then:

• Velocity in positive direction → $v > 0$
• Velocity in negative direction → $v < 0$
• Acceleration in positive direction → $a > 0$
• Acceleration in negative direction → $a < 0$

Problem-Solving Strategy

1. Draw a diagram with coordinate system
2. List known variables ($x_0$, $v_0$, $v$, $a$, $t$)
3. Identify unknown you need to find
4. Choose equation that relates knowns and unknown
5. Solve algebraically before plugging in numbers
6. Check units and reasonableness

Find: Distance traveled ($x$)

Choose equation: We have $v_0$, $a$, and $t$, and need $x$. Missing $v$.

Use: $x = x_0 + v_0 t + \frac{1}{2}at^2$

Solve: $x = 0 + (0)(5.0) + \frac{1}{2}(2.0)(5.0)^2$ $x = 0 + 0 + \frac{1}{2}(2.0)(25)$ $x = \frac{1}{2}(50)$ $x = 25 \text{ m}$

Answer: The car travels 25 meters.

Check: Units are correct (m). A car accelerating for 5 seconds should travel a reasonable distance, and 25 m ≈ 82 feet seems right.