Kinematics in One Dimension

Overview

Kinematics is the study of motion without considering its causes. In one dimension, we analyze motion along a straight line using position, velocity, and acceleration.

Key Quantities

Quantity	Symbol	SI Unit	Description
Position	$x$	m	Location relative to origin
Displacement	$\Delta x$	m	Change in position ($x_2 - x_1$)
Velocity	$v$	m/s	Rate of change of position
Acceleration	$a$	m/s²	Rate of change of velocity
Time	$t$	s	Duration of motion

Average vs Instantaneous

Average Velocity

$v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}$

Instantaneous Velocity

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

Average Acceleration

$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$

Instantaneous Acceleration

$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

Kinematic Equations (Constant Acceleration)

These equations apply when acceleration is constant:

$v = v_0 + at$

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

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

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

$x = x_0 + vt - \frac{1}{2}at^2$

Where:

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

Free Fall

Objects in free fall experience constant acceleration due to gravity:

$g = 9.8 \text{ m/s}^2 \approx 10 \text{ m/s}^2 \quad \text{(downward)}$

For objects thrown upward:

• At maximum height: $v = 0$
• Time to reach max height: $t = v_0/g$
• Maximum height: $h = v_0^2/(2g)$

Sign Conventions

• Choose a positive direction (usually right or up)
• Displacement, velocity, and acceleration are positive in that direction
• Negative values indicate opposite direction

Examples

Example 1: Constant Velocity

A car travels at 25 m/s for 10 seconds. Find the displacement.

$x = v_0 t = (25 \text{ m/s})(10 \text{ s}) = 250 \text{ m}$

Example 2: Constant Acceleration

A car accelerates from rest at 3 m/s² for 8 seconds. Find final velocity and displacement.

$v = v_0 + at = 0 + (3)(8) = 24 \text{ m/s}$

$x = v_0 t + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3)(8)^2 = 96 \text{ m}$

Example 3: Free Fall

A ball is dropped from a height of 45 m. Find the time to hit the ground.

$y = y_0 + v_0 t - \frac{1}{2}gt^2$

$0 = 45 + 0 - \frac{1}{2}(9.8)t^2$

$t^2 = \frac{90}{9.8} = 9.18$

$t = 3.03 \text{ s}$

Problem-Solving Strategy

1. Draw a diagram and choose a coordinate system
2. List known and unknown quantities
3. Select the appropriate kinematic equation(s)
4. Solve for the unknown
5. Check units and reasonableness of answer