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:

• $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i}$

• 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:

• $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i} = \frac{20 - 10}{2 - 0} = 5\ \mathrm{m\,s^{-2}}$

• 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:

• $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i} = \frac{-20 - (-10)}{2 - 0} = \frac{-10}{2} = -5\ \mathrm{m\,s^{-2}}$

• 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: $y(t) = y<em>i + v</em>i t + \tfrac{1}{2} a t^2\qquad (a = -g)$

  • Velocity-position relation: $v^2 = v<em>i^2 + 2 a (y - y</em>i)\qquad (a = -g)$

  • Time to reach the top: t{\text{top}} = \frac{vi}{g}\quad (v_i > 0)

  • Maximum height: $h<em>{\max} = \frac{v</em>i^2}{2 g}$

• 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: $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i}$

• Velocity under constant acceleration: $v(t) = v<em>i + a (t - t</em>i)$

• Position under constant acceleration: $y(t) = y<em>i + v</em>i (t - t<em>i) + \tfrac{1}{2} a (t - t</em>i)^2$

• Energy-like velocity-position relation: $v^2 = v<em>i^2 + 2 a (y - y</em>i)$

• For vertical throw: $a = -g\quad (g \approx 9.81\ \mathrm{m\,s^{-2}})$

• Apex time and height: $t<em>{\text{top}} = \frac{v</em>i}{g}, \quad h<em>{\max} = \frac{v</em>i^2}{2g}$

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:

• $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i}$

• 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: $v<em>i = +10\ \mathrm{m\,s^{-1}}$, $v</em>f = +20\ \mathrm{m\,s^{-1}}$, $t<em>i = 0$, $t</em>f = 2\ \mathrm{s}$

• Calculation:

• $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i} = \frac{20 - 10}{2 - 0} = 5\ \mathrm{m\,s^{-2}}$

• Direction: + (same as the motion).

Example 2: Deceleration with negative velocities

• Given: $v<em>i = -10\ \mathrm{m\,s^{-1}}$, $v</em>f = -20\ \mathrm{m\,s^{-1}}$, $t<em>i = 0$, $t</em>f = 2\ \mathrm{s}$

• Calculation:

• $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i} = \frac{-20 - (-10)}{2 - 0} = \frac{-10}{2} = -5\ \mathrm{m\,s^{-2}}$

• 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: $v<em>i = +10$, $v</em>f = +20$ \rightarrow $a = +5$; velocity and acceleration are in the same (positive) direction; speed increases.

  • Case B: $v<em>i = -10$, $v</em>f = -20$ \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: $y(t) = y<em>i + v</em>i t + \tfrac{1}{2} a t^2\qquad (a = -g)$

  • Velocity-position relation: $v^2 = v<em>i^2 + 2 a (y - y</em>i)\qquad (a = -g)$

  • Time to reach the top: t{\text{top}} = \frac{vi}{g}\quad (v_i > 0)

  • Maximum height: $h<em>{\max} = \frac{v</em>i^2}{2 g}$

• 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: $a = \frac{v<em>f - v</em>i}{t<em>f - t</em>i}$

• Velocity under constant acceleration: $v(t) = v<em>i + a (t - t</em>i)$

• Position under constant acceleration: $y(t) = y<em>i + v</em>i (t - t<em>i) + \tfrac{1}{2} a (t - t</em>i)^2$

• Energy-like velocity-position relation: $v^2 = v<em>i^2 + 2 a (y - y</em>i)$

• For vertical throw: $a = -g\quad (g \approx 9.81\ \mathrm{m\,s^{-2}})$

• Apex time and height: $t<em>{\text{top}} = \frac{v</em>i}{g}, \quad h<em>{\max} = \frac{v</em>i^2}{2g}$