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:

v = v_o - g t
v^2 = v_o^2 - 2 g y
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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