Motion Along a Straight Line: Position, Velocity, and Acceleration

Fundamental Concepts of Position

• Definition: Position is a measurement of a particle's location relative to a specific reference point.
• Reference Point: This is also known as the Origin or Zero. It represents the starting point for any axis measurement ($0$ on the number line).
• Sign Convention and Direction:
  • Positive Direction: Indicated by increasing numbers (e.g., to the right or upward), usually marked with a $+$ sign.
  • Negative Direction: This is the direction opposite to the positive direction, marked with a $-$ sign.
• SI Units of Position: The standard International System unit for position is meters ($m$).
• Visual Representation: On a standard coordinate axis $X(m)$, points to the right of the origin ($0$) are positive ($1, 2, 3$), and points to the left are negative ($-1, -2, -3$).

Displacement

• Definition: A change in position is referred to as displacement, denoted by the symbol $\Delta X$.
• Mathematical Calculation: Displacement is the difference between the final position ($X_2$) and the initial position ($X_1$):     $\Delta X = X_2 - X_1$
• Vector Nature: Displacement is a vector quantity. In one-dimensional motion, its direction is uniquely determined by its sign ($+$ or $-$).
• Units: Like position, the SI unit for displacement is meters ($m$).
• Examples of Displacement:
  1. Positive Displacement: A particle moves from $X_1 = 5\,m$ to $X_2 = 12\,m$.         $\Delta X = 12\,m - 5\,m = 7\,m$ (Positive direction).
  2. Negative Displacement: A particle moves from $X_1 = 5\,m$ to $X_2 = 1\,m$.         $\Delta X = 1\,m - 5\,m = -4\,m$ (Negative direction).
  3. Zero Displacement: A particle moves from $X_1 = 5\,m$ to $X = 200\,m$ and then back to $X_2 = 5\,m$.         $\Delta X = 5\,m - 5\,m = 0\,m$. Note that while the distance traveled is large, the net change in position is zero.

Velocity: Average and Instantaneous

Average Velocity ($v_{avg}$)

• Definition: Average velocity is the ratio of the displacement ($\Delta X$) to the specific time interval ($\Delta t$) during which that displacement occurred.
• Formula:     $v_{avg} = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}$
• SI Units: Meters per second ($m/s$).
• Graphical Interpretation: On a graph of Position ($X$) vs. Time ($t$), the average velocity is the slope of the straight line that connects the two points representing the start and end of the interval.     $\text{Slope} = \frac{\Delta X}{\Delta t}$

Average Speed ($s_{avg}$)

• Definition: Average speed is the ratio of the total distance traveled to the time interval in which that distance was covered.
• Formula:     $s_{avg} = \frac{\text{total distance}}{\Delta t}$
• Key Distinction: Average speed is always positive and does not include a direction (it is a scalar, whereas velocity is a vector).
• Comparative Example: A particle moves from $x = 3\,m$ to $x = -3\,m$ in $2$ seconds.
  • Average Velocity: $v_{avg} = \frac{-3 - 3}{2} = -3\,m/s$.
  • Average Speed: $s_{avg} = \frac{6}{2} = 3\,m/s$ (since total distance is the absolute path length of $6\,m$).

Instantaneous Velocity ($v$)

• Definition: The velocity of an object at a single, specific moment in time.
• Calculus Derivation: It is found by taking the limit of the average velocity as the time interval shrinks toward zero:     $v = \lim_{\Delta t \rightarrow 0} \frac{\Delta X}{\Delta t} = \frac{dx}{dt}$
• Graphical Representation: The slope of the position-time curve ($X$ vs. $t$) at a single instant (the derivative of position).
• Direction of Motion: The sign of the instantaneous velocity represents the direction of motion. The physical direction a particle is moving is always the same as the direction of its velocity.

Acceleration

• Definition: A change in a particle's velocity is defined as acceleration.
• Average Acceleration ($a_{avg}$):     $a_{avg} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$
• Instantaneous Acceleration ($a$):
  • The acceleration at a single moment in time, given by the limit:         $a = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$
  • Second Derivative of Position: Acceleration can also be expressed as the second derivative of position with respect to time:         $a = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^2 x}{dt^2}$
• Graphical Interpretation: Instantaneous acceleration is the slope of the Velocity vs. Time graph.
• Vector Nature and Units:
  • Acceleration is a vector quantity. A positive sign indicates direction along the positive coordinate axis; a negative sign indicates the opposite.
  • Units: Meters per second squared ($m/s^2$).
• NB (Note Well):
  • If the signs of the velocity and acceleration are the same (both positive or both negative), the speed of the particle increases.
  • If the signs of the velocity and acceleration are opposite, the speed of the particle decreases.

Graphical Interpretation of Motion Curves

• Area Under the Acceleration Graph: The change in velocity between two points in time ($t_0$ to $t_1$) is the integral of acceleration over time, which corresponds to the area under the acceleration-time graph:     $v_1 - v_0 = \int_{t_0}^{t_1} a \, dt$
• Area Under the Velocity Graph: The change in position (displacement) between time $t_0$ and $t_1$ is the integral of velocity over time, which corresponds to the area under the velocity-time graph:     $x_1 - x_0 = \int_{t_0}^{t_1} v \, dt$

Constant Acceleration and Kinematic Equations

When acceleration is constant, the average and instantaneous accelerations are equal ($a = a_{avg}$).

Derivation of Primary Equations

1. Velocity as a function of time:     From $a = \frac{v - v_0}{t}$, we rearrange to get:     $v = v_0 + at$

2. Position as a function of time:     Using average velocity for constant acceleration: $v_{avg} = \frac{v + v_0}{2}$.     Substitute this into the displacement formula $x = x_0 + v_{avg}t$:     $x = x_0 + \left( \frac{v_0 + (v_0 + at)}{2} \right) t$$x = x_0 + v_0 t + \frac{1}{2} a t^2$

3. Velocity-Displacement Relationship (Eliminating Time):     Solve the first equation for time: $t = \frac{v - v_0}{a}$.     Substitute this into the position equation to derive:     $v^2 = v_0^2 + 2a(x - x_0)$

Summary of Kinematic Equations

These equations are valid only when acceleration is constant:

• $v = v_0 + at$
• $x = x_0 + v_0 t + \frac{1}{2} a t^2$
• $v^2 = v_0^2 + 2a(x - x_0)$
• $x = x_0 + \frac{1}{2}(v + v_0)t$

Free-Fall Motion

• Definition: Free-fall acceleration is the rate at which an object accelerates downward in the absence of air resistance.
• Standard Value: Near the surface of the Earth, the magnitude of free-fall acceleration is denoted as $g$.
  • $g = +9.81\,m/s^2$
• Characteristics:
  • Acceleration is constant near the surface of the Earth ($a_y = -g$).
  • It is independent of the properties of the object, such as mass, density, or shape.
  • Because it acts downward, it is assigned a negative value in coordinate systems where "up" is positive: $a_y = -9.81\,m/s^2$.

Kinematic Equations for Free-Fall

Substituting $y$ for $x$ and $-g$ for $a$:

• $v_y = v_{0y} - gt$
• $y = y_0 + v_{0y} t - \frac{1}{2} g t^2$
• $v_y^2 = v_{0y}^2 - 2g(y - y_0)$

Practical Examples

Vertical Motion: Ball Thrown from the Ground

Problem: A ball is thrown vertically from the ground to a maximum height of $50\,m$.

Part (a): Find initial speed ($v_{0y}$)

• Given: $y_0 = 0\,m$, $y_f = 50\,m$, $v_{fy} = 0\,m/s$ (at max height), $g = 9.81\,m/s^2$.
• Equation: $v_{fy}^2 = v_{0y}^2 - 2g(y_f - y_0)$
• Calculation:     $0 = v_{0y}^2 - 2(9.81)(50)$$v_{0y} = \sqrt{2 \times 9.81 \times 50} = 31.3\,m/s$

Part (b): Find total time in the air

• Calculation using final velocity at ground ($v_{fy} = -31.3\,m/s$):     $v_{fy} = v_{0y} - gt$$-31.3 = 31.3 - 9.81t$$t = \frac{62.6}{9.81} = 6.38\,s$
• Alternative: Find time to peak ($t_{1/2}$) and double it:     $0 = 31.3 - 9.81 t_{1/2} \rightarrow t_{1/2} = 3.19\,s$$t_{total} = 2 \times 3.19 = 6.38\,s$

Part (c): Graphical Sketches

• Position ($y$) vs. Time: A downward-opening parabola starting at $(0,0)$, peaking at $(3.19, 50)$, and returning to axis at $6.38\,s$.
• Velocity ($v_y$) vs. Time: A straight line with slope $-g$. Intercepts vertical axis at $31.3\,m/s$, crosses time axis at $3.19\,s$, and ends at $-31.3\,m/s$ at $6.38\,s$.
• Acceleration ($a_y$) vs. Time: A flat horizontal line at $-9.81\,m/s^2$.

Motion from an Ascending Balloon

Problem: A package is dropped from a hot air balloon ascending at $12\,m/s$ when it is $80\,m$ above the ground.

Part (a): Time to reach the ground

• Given: $v_{0y} = +12\,m/s$ (package inherits balloon velocity), $y_0 = 80\,m$, $y_f = 0\,m$.
• Equation: $y_f = y_0 + v_{0y}t - \frac{1}{2}gt^2$
• Calculation:     $0 = 80 + 12t - 4.905t^2$
• Using the quadratic formula:     $t = \frac{-12 \pm \sqrt{12^2 - 4(-4.91)(80)}}{2(-4.91)}$$t = 5.44\,s$ (Physical solution) or $t = -2.99\,s$ (Non-physical solution).

Part (b): Impact Speed

• Equation: $v_{fy} = v_{0y} - gt$
• Calculation:     $v_{fy} = 12 - 9.81(5.44) = -41.4\,m/s$
• Speed: Magnitude of velocity, $|v_{fy}| = 41.4\,m/s$.