Acceleration and Kinematics Vocabulary
Acceleration: Core Concepts
Acceleration is the rate of change of velocity; it exists whenever velocity changes with time.
Acceleration is a vector (has both magnitude and direction).
Quick check: if velocity changes over time, there is acceleration; otherwise, a = 0.
Notation and Key Variables
a = acceleration (assumed constant over the interval in the transcript).
v_i = initial velocity.
v_f = final velocity.
t_i = initial time.
t_f = final time.
In the transcript, acceleration is treated as constant over the interval.
Sign Conventions and Vector Nature
Velocity is a vector; its sign (+ or -) denotes direction of motion.
Acceleration is a vector; its sign denotes direction of acceleration.
The direction of motion and the direction of acceleration can be the same or opposite, affecting speed变化.
Key Formula: Constant Acceleration
For constant acceleration over the time interval, the acceleration is:
a = \frac{vf - vi}{tf - ti}
This formula captures the discrete change in velocity over the time interval when acceleration is constant.
Example 1: Car accelerates from 10 m/s to 20 m/s in 2 s
Given: $vi = +10\ \mathrm{m\,s^{-1}}$, $vf = +20\ \mathrm{m\,s^{-1}}$, $ti = 0$, $tf = 2\ \mathrm{s}$
Calculation:
a = \frac{vf - vi}{tf - ti} = \frac{20 - 10}{2 - 0} = 5\ \mathrm{m\,s^{-2}}
Direction: + (same as the motion).
Example 2: Deceleration with negative velocities
Given: $vi = -10\ \mathrm{m\,s^{-1}}$, $vf = -20\ \mathrm{m\,s^{-1}}$, $ti = 0$, $tf = 2\ \mathrm{s}$
Calculation:
a = \frac{vf - vi}{tf - ti} = \frac{-20 - (-10)}{2 - 0} = \frac{-10}{2} = -5\ \mathrm{m\,s^{-2}}
Interpretation: Negative sign indicates the moving direction is opposite to the + direction.
Case: Velocity and acceleration directions
If velocity and acceleration are in the same direction, speed increases.
If velocity and acceleration are in opposite directions, speed decreases.
Examples from the transcript:
Case A: $vi = +10$, $vf = +20$ → $a = +5$; velocity and acceleration are in the same (positive) direction; speed increases.
Case B: $vi = -10$, $vf = -20$ → $a = -5$; velocity and acceleration are in the same (negative) direction; speed increases in the negative direction.
Practical note: If velocity changes sign during the interval, interpret the instantaneous directions carefully; use the given sign conventions consistently.
Ball Thrown Vertically Upward: Sign Convention
Sign conventions (choose upward as positive):
Velocity sign: positive when moving upward, negative when moving downward.
Displacement sign: positive when located above the chosen origin, negative when below.
Release point: often considered positive if it lies in the upward region (UP = +).
Acceleration sign: due to gravity is downward, so $a$ is negative: $a = -g$ with $g \approx 9.81\ \mathrm{m\,s^{-2}}$.
Example states:
Upward displacement increases $y$ (positive), downward motion decreases $y$ (negative displacement).
At launch: $v$ is positive (upward), $a$ is negative (gravity).
At the top: $v = 0$, $a$ remains negative.
During descent: $v$ becomes negative, $a$ remains negative.
Key kinematic relations (under constant gravity):
Velocity-time relation: v(t) = v_i + a t\qquad (a = -g)
Position-time relation: y(t) = yi + vi t + \tfrac{1}{2} a t^2\qquad (a = -g)
Velocity-position relation: v^2 = vi^2 + 2 a (y - yi)\qquad (a = -g)
Time to reach the top: t{\text{top}} = \frac{vi}{g}\quad (v_i > 0)
Maximum height: h{\max} = \frac{vi^2}{2 g}
Quick apex sign intuition: with $v_i > 0$ and gravity downward, the velocity decreases to zero while $y$ increases; after that, $v$ becomes negative and gravity continues to pull downward.
Practical takeaways and sign-aware reasoning
Acceleration exists whenever velocity changes; if velocity is constant, $a = 0$.
Always use a consistent sign convention to avoid sign errors.
When solving problems, pick a positive direction (e.g., +x or +y) and express all quantities with respect to that direction.
The constant-acceleration model is an idealization that works well over short intervals and in many real-world motions.
Real-world relevance and connections
1D motion with constant acceleration is a foundational model for many physical situations, including vehicles, projectiles, and free-fall under gravity.
The sign conventions taught here underpin more advanced topics in physics and engineering, such as work, energy, and impulse where vector directions are critical.
Quick reference formulas (concise)
Acceleration from velocity change: a = \frac{vf - vi}{tf - ti}
Velocity under constant acceleration: v(t) = vi + a (t - ti)
Position under constant acceleration: y(t) = yi + vi (t - ti) + \tfrac{1}{2} a (t - ti)^2
Energy-like velocity-position relation: v^2 = vi^2 + 2 a (y - yi)
For vertical throw: a = -g\quad (g \approx 9.81\ \mathrm{m\,s^{-2}})
Apex time and height: t{\text{top}} = \frac{vi}{g}, \quad h{\max} = \frac{vi^2}{2g}
Acceleration: Core Concepts
Acceleration is the rate of change of velocity; it exists whenever velocity changes with time.
Acceleration is a vector (has both magnitude and direction).
Quick check: if velocity changes over time, there is acceleration; otherwise, a = 0.
Notation and Key Variables
a = acceleration (assumed constant over the interval in the transcript).
v_i = initial velocity.
v_f = final velocity.
t_i = initial time.
t_f = final time.
In the transcript, acceleration is treated as constant over the interval.
Sign Conventions and Vector Nature
Velocity is a vector; its sign (+ or -) denotes direction of motion.
Acceleration is a vector; its sign denotes direction of acceleration.
The direction of motion and the direction of acceleration can be the same or opposite, affecting speed变化.
Key Formula: Constant Acceleration
For constant acceleration over the time interval, the acceleration is:
a = \frac{vf - vi}{tf - ti}
This formula captures the discrete change in velocity over the time interval when acceleration is constant.
Example 1: Car accelerates from 10 m/s to 20 m/s in 2 s
Given: vi = +10\ \mathrm{m\,s^{-1}}, vf = +20\ \mathrm{m\,s^{-1}}, ti = 0, tf = 2\ \mathrm{s}
Calculation:
a = \frac{vf - vi}{tf - ti} = \frac{20 - 10}{2 - 0} = 5\ \mathrm{m\,s^{-2}}
Direction: + (same as the motion).
Example 2: Deceleration with negative velocities
Given: vi = -10\ \mathrm{m\,s^{-1}}, vf = -20\ \mathrm{m\,s^{-1}}, ti = 0, tf = 2\ \mathrm{s}
Calculation:
a = \frac{vf - vi}{tf - ti} = \frac{-20 - (-10)}{2 - 0} = \frac{-10}{2} = -5\ \mathrm{m\,s^{-2}}
Interpretation: Negative sign indicates the moving direction is opposite to the + direction.
Case: Velocity and acceleration directions
If velocity and acceleration are in the same direction, speed increases.
If velocity and acceleration are in opposite directions, speed decreases.
Examples from the transcript:
Case A: vi = +10, vf = +20 \rightarrow a = +5; velocity and acceleration are in the same (positive) direction; speed increases.
Case B: vi = -10, vf = -20 \rightarrow a = -5; velocity and acceleration are in the same (negative) direction; speed increases in the negative direction.
Practical note: If velocity changes sign during the interval, interpret the instantaneous directions carefully; use the given sign conventions consistently.
Ball Thrown Vertically Upward: Sign Convention
Sign conventions (choose upward as positive):
Velocity sign: positive when moving upward, negative when moving downward.
Displacement sign: positive when located above the chosen origin, negative when below.
Release point: often considered positive if it lies in the upward region (UP = +).
Acceleration sign: due to gravity is downward, so a is negative: a = -g with g \approx 9.81\ \mathrm{m\,s^{-2}}.
Example states:
Upward displacement increases y (positive), downward motion decreases y (negative displacement).
At launch: v is positive (upward), a is negative (gravity).
At the top: v = 0, a remains negative.
During descent: v becomes negative, a remains negative.
Key kinematic relations (under constant gravity):
Velocity-time relation: v(t) = v_i + a t\qquad (a = -g)
Position-time relation: y(t) = yi + vi t + \tfrac{1}{2} a t^2\qquad (a = -g)
Velocity-position relation: v^2 = vi^2 + 2 a (y - yi)\qquad (a = -g)
Time to reach the top: t{\text{top}} = \frac{vi}{g}\quad (v_i > 0)
Maximum height: h{\max} = \frac{vi^2}{2 g}
Quick apex sign intuition: with v_i > 0 and gravity downward, the velocity decreases to zero while y increases; after that, v becomes negative and gravity continues to pull downward.
Practical takeaways and sign-aware reasoning
Acceleration exists whenever velocity changes; if velocity is constant, a = 0.
Always use a consistent sign convention to avoid sign errors.
When solving problems, pick a positive direction (e.g., +x or +y) and express all quantities with respect to that direction.
The constant-acceleration model is an idealization that works well over short intervals and in many real-world motions.
Real-world relevance and connections
1D motion with constant acceleration is a foundational model for many physical situations, including vehicles, projectiles, and free-fall under gravity.
The sign conventions taught here underpin more advanced topics in physics and engineering, such as work, energy, and impulse where vector directions are critical.
Quick reference formulas (concise)
Acceleration from velocity change: a = \frac{vf - vi}{tf - ti}
Velocity under constant acceleration: v(t) = vi + a (t - ti)
Position under constant acceleration: y(t) = yi + vi (t - ti) + \tfrac{1}{2} a (t - ti)^2
Energy-like velocity-position relation: v^2 = vi^2 + 2 a (y - yi)
For vertical throw: a = -g\quad (g \approx 9.81\ \mathrm{m\,s^{-2}})
Apex time and height: t{\text{top}} = \frac{vi}{g}, \quad h{\max} = \frac{vi^2}{2g}