Uniformly Accelerated Motion

Definition & Core Characteristics of Uniformly Accelerated Motion (UAM)

• Motion in which an object’s velocity changes by the same amount during every equal time interval.

• Equivalently: acceleration is constant (magnitude and direction remain fixed).

• Consequence: velocity increases or decreases uniformly; displacement–time graph follows a parabola; velocity–time graph is a straight line.

• Everyday intuition: the “push-back” you feel when a jeep or car pulls away from rest is the result of constant forward acceleration while your body’s inertia tries to keep you at rest.

Fundamental Kinematic Equations (1-D, constant a)

• Acceleration from two times:
  

  $a = \frac{v_f - v_o}{t - t_o} \quad (\text{Eq. 1})$

• Final velocity when $a$ and $t$ are known:
  

  $v_f = v_o + a t \quad (\text{Eq. 2})$

• Displacement from average velocity:
  

  $d = \frac{v_f + v_o}{2}\,t \quad (\text{Eq. 3})$

• Displacement from $a$ and $t$ (no $v_f$): $d = v_o t + \frac{1}{2} a t^2 \quad (\text{Eq. 4})$

• Relation that removes $t$ (no explicit time):
  

  $v_f^2 = v_o^2 + 2 a d \quad (\text{Eq. 5})$

Symbols, Meanings & SI Units

• $v_o$ – initial velocity (m/s)

• $v_f$ – final velocity (m/s)

• $a$ or $\alpha$ – acceleration (m/s$^2$)

• $t$ – elapsed time (s)

• $d$ (or $x,\;y$) – displacement (m)

Worked Examples in the Transcript

• Problem 1 – Motorcycle acceleration

• Given $v_o = 10\,\text{m/s},\; v_f = 30\,\text{m/s},\; t = 5\,\text{s}$

• $a = \tfrac{30-10}{5} = 4\,\text{m/s}^2$
  • Problem 2 – Train leaving station

• $v_o = 0,\; a = 3\,\text{m/s}^2,\; t = 8\,\text{s}$

• $v_f = 0 + 3(8) = 24\,\text{m/s}$
  • Problem 3 – Car’s travel distance during speed-up

• $v_o = 10\,\text{m/s},\; v_f = 30\,\text{m/s},\; t = 4\,\text{s}$

• $d = \tfrac{10+30}{2}(4) = 80\,\text{m}$
  • Problem 4 – Scooter distance via $v_f^2$ relation

• $v_o = 5\,\text{m/s},\; v_f = 25\,\text{m/s},\; a = 5\,\text{m/s}^2$

• $d = \tfrac{25^2 - 5^2}{2(5)} = 60\,\text{m}$
  • Problem 5 – Bus from rest

• $v_o = 0,\; a = 2\,\text{m/s}^2,\; t = 6\,\text{s}$

• $d = 0(6) + \tfrac{1}{2}(2)(6)^2 = 36\,\text{m}$

Free-Fall as a Special Case of UAM (Vertical Axis)

• Gravity supplies a nearly constant acceleration $g \approx 9.8\,\text{m/s}^2$ downward.

• Sign convention: choose upward as positive. Then $a = -g$.

• Kinematic set becomes:

1. $v = v_o - g t$

2. $v^2 = v_o^2 - 2 g y$

3. $y = v_o t - \tfrac{1}{2} g t^2$

• Caution: Insert the negative sign once—do not also substitute $g = -9.8$ afterwards.

Example – Stone dropped from rest

• Data: $v_o = 0,\; g = 9.8\,\text{m/s}^2$

• After 1 s:

  • Velocity: $v = 0 - 9.8(1) = -9.8\,\text{m/s}$ (downward)

  • Displacement: $y = 0(1) - \tfrac{1}{2}(9.8)(1)^2 = -4.9\,\text{m}$ (4.9 m below release point)

Example – Ball thrown upward at $29.4\,\text{m/s}$

• Velocity each second via $v = 29.4 - 9.8 t$

  • $t = 1\,\text{s}:\; v = 19.6\,\text{m/s}$ upward

  • $t = 2\,\text{s}:\; v = 9.8\,\text{m/s}$ upward

  • $t = 3\,\text{s}:\; v = 0\,\text{m/s}$ (peak)

• Continues downward with negative values thereafter.

• Demonstration note: In a vacuum, feathers and stones fall together—air resistance is the only reason light objects normally lag.

Graphical Interpretation of UAM

• Zero acceleration: $d \text{ vs } t$ is a straight line; $v \text{ vs } t$ is a horizontal line.

• Constant positive $a$:

  • $d \text{ vs } t$: upward-opening parabola.

  • $v \text{ vs } t$: straight line with positive slope.

  • $a \text{ vs } t$: horizontal line at $+a$.

Learning Objectives Stated in Module 4

• Solve for unknowns in one-dimensional UAM equations.

• Treat $g = 9.8\,\text{m/s}^2$ as constant at Earth’s surface.

• Apply UAM to tailgating, pursuit/chase, rocket launches, free-fall, and similar contexts.

Additional Physical Relations & Constants Mentioned (Minor References)

• Newton’s 2nd law: $F = m a$.

• Density: $\rho = \dfrac{m}{V}$.

• Ideal-gas law: PV = n R T\,\,(R \approx 8.314\,\text{J/(mol·K)}).

• Boltzmann constant: $k_B \approx 1.38 \times 10^{-23}\,\text{J/K}$.

• Energy–mass equivalence: $E = m c^2$.

• Ohm’s law (written as $I = \tfrac{U}{R}$ in transcript, equivalent to $I = \tfrac{V}{R}$).

• Unit conversion: $1\,\text{m} = 1000\,\text{mm}$; $1\,\text{atm} = 101325\,\text{Pa}$.

Problem-

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Definition & Core Characteristics of Uniformly Accelerated Motion (UAM)

• Motion in which an object’s velocity changes by the same amount during every equal time interval.

• Equivalently: acceleration is constant (magnitude and direction remain fixed).

• Consequence: velocity increases or decreases uniformly; displacement–time graph follows a parabola; velocity–time graph is a straight line.

• Everyday intuition: the “push-back” you feel when a jeep or car pulls away from rest is the result of constant forward acceleration while your body’s inertia tries to keep you at rest.

Fundamental Kinematic Equations (1-D, constant a)

• Acceleration from two times:
  

  $a = \frac{v_f - v_o}{t - t_o} \quad (\text{Eq. 1})$

• Final velocity when $a$ and $t$ are known:
  

  $v_f = v_o + a t \quad (\text{Eq. 2})$

• Displacement from average velocity:
  

  $d = \frac{v_f + v_o}{2}\,t \quad (\text{Eq. 3})$

• Displacement from $a$ and $t$ (no $v_f$): $d = v_o t + \frac{1}{2} a t^2 \quad (\text{Eq. 4})$

• Relation that removes $t$ (no explicit time):
  

  $v_f^2 = v_o^2 + 2 a d \quad (\text{Eq. 5})$

Symbols, Meanings & SI Units

• $v_o$ – initial velocity (m/s)

• $v_f$ – final velocity (m/s)

• $a$ or $\alpha$ – acceleration (m/s$^2$)

• $t$ – elapsed time (s)

• $d$ (or $x,\;y$) – displacement (m)

Worked Examples in the Transcript

• Problem 1 – Motorcycle acceleration

• Given $v_o = 10\,\text{m/s},\; v_f = 30\,\text{m/s},\; t = 5\,\text{s}$

• $a = \tfrac{30-10}{5} = 4\,\text{m/s}^2$

• Problem 2 – Train leaving station

• $v_o = 0,\; a = 3\,\text{m/s}^2,\; t = 8\,\text{s}$

• $v_f = 0 + 3(8) = 24\,\text{m/s}$

• Problem 3 – Car’s travel distance during speed-up

• $v_o = 10\,\text{m/s},\; v_f = 30\,\text{m/s},\; t = 4\,\text{s}$

• $d = \tfrac{10+30}{2}(4) = 80\,\text{m}$

• Problem 4 – Scooter distance via $v_f^2$ relation

• $v_o = 5\,\text{m/s},\; v_f = 25\,\text{m/s},\; a = 5\,\text{m/s}^2$

• $d = \tfrac{25^2 - 5^2}{2(5)} = 60\,\text{m}$

• Problem 5 – Bus from rest

• $v_o = 0,\; a = 2\,\text{m/s}^2,\; t = 6\,\text{s}$

• $d = 0(6) + \tfrac{1}{2}(2)(6)^2 = 36\,\text{m}$

Free-Fall as a Special Case of UAM (Vertical Axis)

• Gravity supplies a nearly constant acceleration $g \approx 9.8\,\text{m/s}^2$ downward.

• Sign convention: choose upward as positive. Then $a = -g$.

• Kinematic set becomes:

1. $v = v_o - g t$

2. $v^2 = v_o^2 - 2 g y$

3. $y = v_o t - \tfrac{1}{2} g t^2$

• Caution: Insert the negative sign once—do not also substitute $g = -9.8$ afterwards.

Example – Stone dropped from rest

• Data: $v_o = 0,\; g = 9.8\,\text{m/s}^2$

• After 1 s:

  • Velocity: $v = 0 - 9.8(1) = -9.8\,\text{m/s}$ (downward)

  • Displacement: $y = 0(1) - \tfrac{1}{2}(9.8)(1)^2 = -4.9\,\text{m}$ (4.9 m below release point)

Example – Ball thrown upward at $29.4\,\text{m/s}$

• Velocity each second via $v = 29.4 - 9.8 t$

  • $t = 1\,\text{s}:\; v = 19.6\,\text{m/s}$ upward

  • $t = 2\,\text{s}:\; v = 9.8\,\text{m/s}$ upward

  • $t = 3\,\text{s}:\; v = 0\,\text{m/s}$ (peak)

• Continues downward with negative values thereafter.

• Demonstration note: In a vacuum, feathers and stones fall together—air resistance is the only reason light objects normally lag.

Graphical Interpretation of UAM

• Zero acceleration: $d \text{ vs } t$ is a straight line; $v \text{ vs } t$ is a horizontal line.

• Constant positive $a$:

  • $d \text{ vs } t$: upward-opening parabola.

  • $v \text{ vs } t$: straight line with positive slope.

  • $a \text{ vs } t$: horizontal line at $+a$.

Learning Objectives Stated in Module 4

• Solve for unknowns in one-dimensional UAM equations.

• Treat $g = 9.8\,\text{m/s}^2$ as constant at Earth’s surface.

• Apply UAM to tailgating, pursuit/chase, rocket launches, free-fall, and similar contexts.

Additional Physical Relations & Constants Mentioned (Minor References)

• Newton’s 2nd law: $F = m a$.

• Density: $\rho = \dfrac{m}{V}$.

• Ideal-gas law: PV = n R T\,\,(R \approx 8.314\,\text{J/(mol·K)}).

• Boltzmann constant: $k_B \approx 1.38 \times 10^{-23}\,\text{J/K}$.

• Energy–mass equivalence: $E = m c^2$.

• Ohm’s law (written as $I = \tfrac{U}{R}$