Comprehensive Notes on Motion Along a Straight Line

Displacement and One-Dimensional Coordinate Systems

Motion along a straight line, typically designated as the $x$ axis, is defined by the displacement of a particle from an initial position $x_i$ to a final position $x_f$. The displacement, denoted as $\Delta x$, is calculated as $\Delta x = x_f - x_i$. The magnitude of displacement is the absolute value of this difference, $|\Delta x|$, whereas the sign of the displacement indicates the direction of motion relative to the origin. For example, if a particle moves from $x_i = -4\,m$ to $x_f = 4\,m$, the displacement is $8\,m$, which represents a larger magnitude than moving from $x_i = 4\,m$ to $x_f = 6\,m$ ($2\,m$) or from $x_i = -4\,m$ to $x_f = -8\,m$ (magnitude of $4\,m$). Negative displacement occurs whenever the final coordinate is less than the initial coordinate, such as moving from $x_i = -4\,m$ to $x_f = -8\,m$, resulting in $\Delta x = -4\,m$.

Average Speed and Velocity in Linear Motion

Average speed is a scalar quantity defined as the total distance covered during a specific time interval divided by that time interval. It does not account for direction and is not necessarily equal to the magnitude of the average velocity. Average velocity is a vector quantity defined as the displacement divided by the time interval: $v_{avg} = \frac{\Delta x}{\Delta t}$. In a round trip scenario, such as a car traveling $50\,km$ from a point called "Hither" to "Yon" and immediately returning to "Hither" over a total of $2\,hours$, the total displacement is $0\,km$, making the magnitude of the average velocity $0\,km/h$. However, the total distance covered is $100\,km$, resulting in an average speed of $50\,km/h$.

For multi-stage trips with varying speeds, the average speed must be calculated using the total distance and total time. For instance, if a car travels $40\,km$ at $80\,km/h$ (taking $0.5\,hours$) and then another $40\,km$ at $40\,km/h$ (taking $1.0\,hour$), the total distance is $80\,km$ and the total time is $1.5\,hours$. The average speed for the entire trip is $\frac{80\,km}{1.5\,h} \approx 53.3\,km/h$.

Relative Motion of Objects

When two objects move toward each other, their relative speed is the sum of their individual speeds. If two automobiles are initially separated by $150\,kilometers$ and are traveling toward one another at speeds of $60\,km/h$ and $40\,km/h$ respectively, their closing speed is $100\,km/h$. The time ($t$) required for them to meet is calculated by dividing the initial separation by the relative speed: $t = \frac{150\,km}{100\,km/h} = 1.5\,hours$.

Calculus and Kinematics

Velocity and acceleration can be derived from the position function $x(t)$ using calculus. The instantaneous velocity $v(t)$ is the first derivative of position with respect to time, $\frac{dx}{dt}$, and the instantaneous acceleration $a(t)$ is the derivative of velocity, $\frac{dv}{dt}$, or the second derivative of position, $\frac{d^2x}{dt^2}$.

Consider a particle with a coordinate given by $x(t) = 16t - 3.0t^3$. To find when the particle is momentarily at rest, one must set the velocity equation to zero: $v(t) = 16 - 9.0t^2 = 0$. Solving for $t$ gives $t = \sqrt{\frac{16}{9}} \approx 1.3\,s$. Similarly, the acceleration at a specific time can be found. For the function $x(t) = 27t - 4.0t^3$, the velocity is $v(t) = 27 - 12t^2$ and the acceleration is $a(t) = -24t$. At $t = 1.0\,s$, the acceleration is $-24\,m/s^2$.

Distance can also be found by integrating the velocity function. If a drag racing car moves with velocity $v = bt^2$, the distance traveled from $t = 0$ is the integral $\int_0^t bt^2 \,dt = \frac{bt^3}{3}$.

Constant Acceleration and Kinematic Equations

Constant acceleration implies that the velocity of an object changes by the same amount every second. For an object with constant acceleration, the following relationships hold:

1. $v = v_0 + at$
2. $x = x_0 + v_0t + \frac{1}{2}at^2$
3. $v^2 = v_0^2 + 2a(x - x_0)$
4. $v_{avg} = \frac{v_0 + v}{2}$

Identification of constant acceleration from position-time functions requires that the highest power of $t$ be no greater than $2$. For example, a particle with $x(t) = 3.5 + 2.7t^2$ has a constant acceleration of $5.4\,m/s^2$, whereas a function with $t^3$ represents variable acceleration.

A car accelerating from rest ($v_0 = 0$) with $a = 4\,m/s^2$ covers distance from $x = 2\,m$ to $x = 8\,m$. Using $v^2 = 2ax$, the velocity at $x = 2\,m$ is $4\,m/s$ and at $x = 8\,m$ is $8\,m/s$. The average velocity during this specific displacement is the mean of the two: $\frac{4 + 8}{2} = 6\,m/s$.

Free Fall and Gravity

Free fall occurs when an object moves solely under the influence of gravity. The acceleration due to gravity, denoted as $g$, is approximately $9.8\,m/s^2$ (or $10\,m/s^2$ for simplified calculations) and is directed downward at all times during flight, whether the object is ascending, descending, or at its highest point.

Key behaviors in free fall include:

• At the highest point of a vertical trajectory, the instantaneous velocity is zero, but the acceleration is still $g$ downward.
• If an object is released from rest, its speed increases by $9.8\,m/s$ every second. Its average velocity during the first second of fall is $4.9\,m/s$, and it falls a distance of $4.9\,m$ during that first second.
• Distance fallen during specific intervals: An object falls further in subsequent seconds because its velocity is increasing. To find the distance fallen during the second second of fall, calculate the position at $t = 2\,s$ ($19.6\,m$) and subtract the position at $t = 1\,s$ ($4.9\,m$), yielding $14.7\,m \approx 15\,m$.
• Proportionality: The maximum height $H$ reached by a projectile is proportional to the square of its initial velocity, $H = \frac{v_0^2}{2g}$. If the initial velocity is doubled, the maximum height increases by a factor of four ($2^2$). If the gravity of a world is doubled ($19.6\,m/s^2$), the maximum height reached for the same initial velocity is halved.

Graphical Analysis of Motion

Graphs provide crucial visual data regarding motion:

• Coordinate vs. Time ($x-t$): The slope of this graph represents the instantaneous velocity. A straight line indicates constant velocity. A curved line (parabola) indicates constant acceleration. If the slope increases, the object is speeding up.
• Velocity vs. Time ($v-t$): The slope of this graph represents the acceleration. The area under the curve represents the displacement $\Delta x$. A horizontal line indicates constant velocity ($a = 0$). If the graph crosses the $t$-axis, the object has reversed direction.
• Acceleration vs. Time ($a-t$): The area under the curve represents the change in velocity $\Delta v$. For an object moving at a constant velocity, the acceleration graph is a horizontal line at $a = 0$.

Multiple Choice Practice and Discussion

1. Displacement magnitude comparison: Given coordinates $x_i$, $x_f$. The result $x_i = -4\,m, x_f = 4\,m$ provides the largest displacement magnitude ($8\,m$).

2. Identifying negative displacement: Moving from $x_i = -4\,m$ to $x_f = -8\,m$ results in $\Delta x = -4\,m$.

3. Average speed definition: Always the distance covered divided by the time interval.

4. Car Pursuit: A car at $3\,m/s^2$ from rest chasing a truck at a constant $15\,m/s$. Setting positions equal: $15t = 0.5(3)t^2 \rightarrow 15 = 1.5t \rightarrow t = 10\,s$.

5. Acceleration vector direction: In free fall (upward is positive), displacement is positive during ascent and negative during descent, but acceleration is negative (downward) at all times.

6. Terminal constraints: It is impossible for a body to have constant velocity and variable acceleration simultaneously, as constant velocity implies zero acceleration.

7. Speeding up criteria: A particle is speeding up if its velocity and acceleration share the same sign (both positive or both negative). For example, if velocity is negative and acceleration is negative, the speed is increasing.

8. Free Fall from a moving platform: If a rocket accelerating upward at $9.8\,m/s^2$ releases a projectile, the projectile’s acceleration immediately becomes $9.8\,m/s^2$ downward due to gravity, despite its initial upward velocity inherited from the rocket.

9. Max height ratios: An object thrown at $100\,m/s$ reaches a maximum height $100$ times higher than one thrown at $10\,m/s$ because $\frac{100^2}{10^2} = \frac{10000}{100} = 100$.

10. Velocity-time graph interpretation: At a point where the graph crosses the time axis from positive to negative velocity, the car is traveling in the reverse direction relative to its earlier movement at positive velocity points.