Motion in a Straight Line Study Guide

Introduction to Motion in a Straight Line

• Definition of Motion:

  • Motion is defined as the change in position of an object with time.

  • Motion is universal across all scales in nature:

  • Everyday human actions: walking, running, riding a bicycle.

  • Physiological processes: movement of air into and out of lungs during sleep, blood flow through arteries and veins.

  • Environmental phenomena: leaves falling from trees, water flowing down a dam.

  • Transportation: movement of automobiles and airplanes.

  • Astronomical phenomena: Earth rotates once every $24\,\text{hours}$ and revolves around the Sun once per year; the Sun moves within the Milky Way galaxy, which itself moves within its local group of galaxies.

• Scope and Approximations:

  • Rectilinear Motion: The study of motion along a straight line.

  • Point Object Approximation: An object in motion is treated as a point object if its physical size is much smaller than the distance it traverses in a reasonable duration of time. In these scenarios, the object's dimensions can be neglected without introducing significant error.

  • Kinematics: The branch of physics focused on describing motion without analyzing the forces or causes that produce it.

Instantaneous Velocity and Speed

• Distinction Between Average and Instantaneous Velocity:

  • Average velocity provides the overall rate of position change over a finite time interval, but fails to indicate how fast an object moves at specific instants within that interval.

  • Instantaneous Velocity ($v$): The velocity of an object at a specific instant of time $t$. It is defined mathematically as the limit of average velocity as the time interval $\Delta t$ approaches zero:     $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

  • The expression $\frac{dx}{dt}$ represents the differential coefficient of position $x$ with respect to time $t$, indicating the instantaneous rate of change of position.

• Methods of Determination:

  • Graphical Method:

  • On a position-time ($x-t$) graph, the instantaneous velocity at time $t$ is given by the slope of the tangent line drawn to the curve at that specific point $P$

• Numerical Method (Limiting Process):

  • For an object moving according to $x = 0.08\,t^3$, the velocity at $t = 4.0\,\text{s}$ is calculated by evaluating $\frac{\Delta x}{\Delta t}$ over progressively smaller time intervals $\Delta t$ centered at $t = 4.0\,\text{s}$, using $t_1 = t - \frac{\Delta t}{2}$ and $t_2 = t + \frac{\Delta t}{2}$:

    • For $\Delta t = 2.0\,\text{s}$ ($t_1 = 3.0\,\text{s}, t_2 = 5.0\,\text{s}$): $x(t_1) = 2.16\,\text{m}$, $x(t_2) = 10.0\,\text{m}$, $\Delta x = 7.84\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.92\,\text{m\,s}^{-1}$.

    • For $\Delta t = 1.0\,\text{s}$ ($t_1 = 3.5\,\text{s}, t_2 = 4.5\,\text{s}$): $x(t_1) = 3.43\,\text{m}$, $x(t_2) = 7.29\,\text{m}$, $\Delta x = 3.86\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.86\,\text{m\,s}^{-1}$.

    • For $\Delta t = 0.5\,\text{s}$ ($t_1 = 3.75\,\text{s}, t_2 = 4.25\,\text{s}$): $x(t_1) = 4.21875\,\text{m}$, $x(t_2) = 6.14125\,\text{m}$, $\Delta x = 1.9225\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.845\,\text{m\,s}^{-1}$.

    • For $\Delta t = 0.1\,\text{s}$ ($t_1 = 3.95\,\text{s}, t_2 = 4.05\,\text{s}$): $x(t_1) = 4.93039\,\text{m}$, $x(t_2) = 5.31441\,\text{m}$, $\Delta x = 0.38402\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.8402\,\text{m\,s}^{-1}$.

    • For $\Delta t = 0.01\,\text{s}$ ($t_1 = 3.995\,\text{s}, t_2 = 4.005\,\text{s}$): $x(t_1) = 5.100824\,\text{m}$, $x(t_2) = 5.139224\,\text{m}$, $\Delta x = 0.0384\,\text{m}$, $\frac{\Delta x}{\Delta t} = 3.8400\,\text{m\,s}^{-1}$.

  • As $\Delta t$ decreases from $2.0\,\text{s}$ to $0.010\,\text{s}$, the average velocity approaches a limiting value of $3.84\,\text{m\,s}^{-1}$, which represents the instantaneous velocity $\frac{dx}{dt}$ at $t = 4.0\,\text{s}$.

  • Analytical Calculation Example:

• For an object moving along the x-axis with position given by $x = a + b\,t^2$, where $a = 8.5\,\text{m}$, $b = 2.5\,\text{m\,s}^{-2}$, and $t$ is in seconds:

  • Velocity equation:       $v = \frac{dx}{dt} = \frac{d}{dt}(a + b\,t^2) = 2b\,t = 5.0\,t\,\text{m\,s}^{-1}$

  • At $t = 0\,\text{s}$: $v = 0\,\text{m\,s}^{-1}$.

  • At $t = 2.0\,\text{s}$: $v = 10\,\text{m\,s}^{-1}$.

  • Average velocity between $t = 2.0\,\text{s}$ and $t = 4.0\,\text{s}$:       $\text{Average velocity} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{(a + 16b) - (a + 4b)}{2.0} = 6.0 \times b = 6.0 \times 2.5 = 15\,\text{m\,s}^{-1}$

  • Properties of Uniform Motion and Instantaneous Speed:

• In uniform motion, velocity remains constant at all times, making instantaneous velocity equal to average velocity at every instant.

• Instantaneous Speed: The magnitude of instantaneous velocity.

  • Example: Velocities of $+24.0\,\text{m\,s}^{-1}$ and $-24.0\,\text{m\,s}^{-1}$ both correspond to an instantaneous speed of $24.0\,\text{m\,s}^{-1}$.

  • While average speed over a finite interval can be greater than or equal to the magnitude of average velocity, instantaneous speed at any instant is strictly equal to the magnitude of instantaneous velocity at that exact instant.

Acceleration

• Development of Concept:

  • Galileo established through studies of freely falling bodies and objects on inclined planes that the rate of velocity change with time is a constant of motion for free fall.

  • Rate of change of velocity with distance decreases as fall distance increases; hence, acceleration is defined as the rate of change of velocity with respect to time.

• Average Acceleration:

  • Ratio of velocity change to the corresponding time interval:     $\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$

  • SI Unit: $\text{m\,s}^{-2}$.

  • Graphically represented as the slope of the straight line joining points $(v_1, t_1)$ and $(v_2, t_2)$ on a velocity-time ($v-t$) graph.

• Instantaneous Acceleration:

  • Limit of average acceleration as $\Delta t$ approaches zero:     $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$

  • Graphically represented as the slope of the tangent line to the $v-t$ curve at that instant.

  • Because velocity is a vector quantity (having magnitude and direction), acceleration can result from a change in speed, a change in direction, or changes in both.

• Graphical Patterns:

  • Position-Time ($x-t$) Graphs:

  • Positive acceleration: curves upward.

  • Negative acceleration: curves downward.

  • Zero acceleration: straight line.

  • Velocity-Time ($v-t$) Graphs for Constant Acceleration:

  • Motion in positive direction with positive acceleration.

  • Motion in positive direction with negative acceleration.

  • Motion in negative direction with negative acceleration.

  • Motion changing direction at time $t_1$ (moves in positive x-direction from $0$ to $t_1$, then turns back in negative direction between $t_1$ and $t_2$).

Area Under Velocity-Time Curve

• Displacement Significance:

  • The area under the velocity-time curve over a given time interval equals the displacement of the object during that interval.

  • For constant velocity $u$ between $t = 0$ and $t = T$:     $\text{Area} = u \times T = u\,T = \text{displacement}$

  • Sharp kinks in $x-t$, $v-t$, and $a-t$ curves imply non-differentiable points. In realistic physical scenarios, graphs are smooth and differentiable everywhere, meaning acceleration and velocity change continuously rather than abruptly.

Kinematic Equations for Uniformly Accelerated Motion

• Derivation from Velocity-Time Curve:

  • From a $v-t$ graph with initial velocity $v_0$ at $t = 0$ and final velocity $v$ at time $t$:     $v = v_0 + a\,t$

  • Total area under $v-t$ curve between $0$ and $t$ equals Area of triangle $ABC$ + Area of rectangle $OACD$:     $x = \frac{1}{2}(v - v_0)t + v_0\,t$

  • Substituting $v - v_0 = a\,t$ gives:     $x = v_0\,t + \frac{1}{2}a\,t^2$

  • Expressing displacement using average velocity $\bar{v}$:     $x = \bar{v}\,t = \left(\frac{v + v_0}{2}\right)t$

  • Substituting $t = \frac{v - v_0}{a}$:     $x = \left(\frac{v + v_0}{2}\right)\left(\frac{v - v_0}{a}\right) = \frac{v^2 - v_0^2}{2a} \implies v^2 = v_0^2 + 2a\,x$

• Set of Kinematic Equations:

  • For origin at initial position $x_0 = 0$:

  1. $v = v_0 + a\,t$

  2. $x = v_0\,t + \frac{1}{2}a\,t^2$

  3. $v^2 = v_0^2 + 2a\,x$

  • For general initial position $x_0 \neq 0$ at $t = 0$:

  1. $v = v_0 + a\,t$

  2. $x = x_0 + v_0\,t + \frac{1}{2}a\,t^2$

  3. $v^2 = v_0^2 + 2a(x - x_0)$

• Calculus Method Derivation:

  • Integrating acceleration:     $a = \frac{dv}{dt} \implies dv = a\,dt$     $\int_{v_0}^v dv = \int_0^t a\,dt = a \int_0^t dt \implies v - v_0 = a\,t \implies v = v_0 + a\,t$

  • Integrating velocity:     $v = \frac{dx}{dt} \implies dx = v\,dt = (v_0 + a\,t)dt$     $\int_{x_0}^x dx = \int_0^t (v_0 + a\,t)dt \implies x - x_0 = v_0\,t + \frac{1}{2}a\,t^2$

  • Acceleration as a function of position:     $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} \implies v\,dv = a\,dx$     $\int_{v_0}^v v\,dv = \int_{x_0}^x a\,dx \implies \frac{v^2 - v_0^2}{2} = a(x - x_0) \implies v^2 = v_0^2 + 2a(x - x_0)$

  • The calculus method possesses the unique advantage of being applicable to non-uniform acceleration as well.

Worked Examples and Applications

• Vertical Motion under Gravity:

  • A ball is thrown vertically upward with initial velocity $20\,\text{m\,s}^{-1}$ from a building height of $25.0\,\text{m}$ ($g = 10\,\text{m\,s}^{-2}$).

  • Part (a): Maximum Height Reached:

  • Taking y-axis vertically upward with zero at ground level ($y_0 = 25.0\,\text{m}$), at maximum height $v = 0\,\text{m\,s}^{-1}$ and $a = -g = -10\,\text{m\,s}^{-2}$.

  • $v^2 = v_0^2 + 2a(y - y_0) \implies 0 = (20)^2 + 2(-10)(y - y_0) \implies y - y_0 = 20\,\text{m}$.

  • Part (b): Total Time to Hit Ground:

  • First Method (Split Path):

    • Upward motion ($A \to B$): $v = v_0 + a\,t_1 \implies 0 = 20 - 10\,t_1 \implies t_1 = 2\,\text{s}$.

    • Downward motion ($B \to C$): Maximum height $y_0 = 45\,\text{m}$, final position $y = 0\,\text{m}$, initial velocity at peak $v_0 = 0\,\text{m\,s}^{-1}$, $a = -10\,\text{m\,s}^{-2}$.

    • $y = y_0 + v_0\,t_2 + \frac{1}{2}a\,t_2^2 \implies 0 = 45 + \frac{1}{2}(-10)t_2^2 \implies 5\,t_2^2 = 45 \implies t_2 = 3\,\text{s}$.

    • Total time = $t_1 + t_2 = 2\,\text{s} + 3\,\text{s} = 5\,\text{s}$.

  • Second Method (Direct Coordinate Method):

    • Initial coordinate $y_0 = 25\,\text{m}$, final coordinate $y = 0\,\text{m}$, $v_0 = 20\,\text{m\,s}^{-1}$, $a = -10\,\text{m\,s}^{-2}$.

    • $y = y_0 + v_0\,t + \frac{1}{2}a\,t^2 \implies 0 = 25 + 20\,t + \frac{1}{2}(-10)t^2$

    • $5\,t^2 - 20\,t - 25 = 0 \implies t^2 - 4\,t - 5 = 0 \implies t = 5\,\text{s}$.

• Free Fall Motion:

  • Free fall refers to the motion of an object released near Earth's surface accelerated downward solely by gravity, neglecting air resistance.

  • For small fall heights relative to Earth's radius, $g = 9.8\,\text{m\,s}^{-2}$ (constant uniform acceleration).

  • Choosing upward direction as positive ($a = -g = -9.8\,\text{m\,s}^{-2}$) with release from rest at $y = 0$ ($v_0 = 0$):     $v = -g\,t = -9.8\,t\,\text{m\,s}^{-1}$     $y = -\frac{1}{2}g\,t^2 = -4.9\,t^2\,\text{m}$     $v^2 = -2g\,y = -19.6\,y\,\text{m}^2\text{s}^{-2}$

• Galileo's Law of Odd Numbers:

  • Statement: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1:3:5:7…]."

  • Proof:

  • Distance from rest is $y = -\frac{1}{2}g\,t^2$. Let $y_0 = -\frac{1}{2}g\,\tau^2$.

  • At $t = 0$: position $y = 0$, distance traversed = $0$.

  • At $t = \tau$: position $y = y_0$, distance in interval = $y_0$, ratio = $1$.

  • At $t = 2\tau$: position $y = 4y_0$, distance in interval = $3y_0$, ratio = $3$.

  • At $t = 3\tau$: position $y = 9y_0$, distance in interval = $5y_0$, ratio = $5$.

  • At $t = 4\tau$: position $y = 16y_0$, distance in interval = $7y_0$, ratio = $7$.

  • At $t = 5\tau$: position $y = 25y_0$, distance in interval = $9y_0$, ratio = $9$.

  • At $t = 6\tau$: position $y = 36y_0$, distance in interval = $11y_0$, ratio = $11$.

  • Galileo Galilei (1564–1642) established this law during the first quantitative studies of free fall.

• Stopping Distance of Vehicles:

  • Definition: The distance a vehicle travels before stopping after brakes are applied.

  • Derivation: From $v^2 = v_0^2 + 2a\,x$, setting final velocity $v = 0$, deceleration $-a$, and stopping distance $d_s$:     $0 = v_0^2 - 2a\,d_s \implies d_s = \frac{v_0^2}{2a}$

  • Stopping distance is proportional to the square of initial velocity ($d_s \propto v_0^2$). Doubling initial velocity increases stopping distance by a factor of 4 for equal deceleration.

  • Empirical braking distance data for a specific car model:

  • Speed $11\,\text{m\,s}^{-1} \implies$ Stopping distance $10\,\text{m}$

  • Speed $15\,\text{m\,s}^{-1} \implies$ Stopping distance $20\,\text{m}$

  • Speed $20\,\text{m\,s}^{-1} \implies$ Stopping distance $34\,\text{m}$

  • Speed $25\,\text{m\,s}^{-1} \implies$ Stopping distance $50\,\text{m}$

• Reaction Time:

  • Definition: The time interval elapsed between observing a situation demanding action, thinking, and executing the action.

  • Measurement via Ruler Drop Experiment:

  • A ruler held vertically is dropped by a friend and caught between the thumb and forefinger.

- Ruler falls freely from rest ($v_0 = 0$, $a = g = 9.8\,\text{m\,s}^{-2}$). Distance $d$ and reaction time $t_r$ are related by:

      $d = \frac{1}{2}g\,t_r^2 \implies t_r = \sqrt{\frac{2d}{g}}$ - For measured drop distance $d = 21.0\,\text{cm} = 0.21\,\text{m}$:       $t_r = \sqrt{\frac{2 \times 0.21}{9.8}} \approx 0.2\,\text{s}$

Summary of Kinematic Quantities

• Path Length:

  • Dimension: $[\text{L}]$

  • Unit: $\text{m}$

  • Remarks: Total path traversed by an object.

• Displacement:

  • Symbol: $\Delta x$

  • Dimension: $[\text{L}]$

  • Unit: $\text{m}$

  • Formula / Remarks: $\Delta x = x_2 - x_1$. In one-dimensional motion, sign indicates direction.

• Average Velocity:

  • Symbol: $\bar{v}$

  • Dimension: $[\text{LT}^{-1}]$

  • Unit: $\text{m\,s}^{-1}$

  • Formula / Remarks: $\bar{v} = \frac{\Delta x}{\Delta t}$.

• Instantaneous Velocity:

  • Symbol: $v$

  • Dimension: $[\text{LT}^{-1}]$

  • Unit: $\text{m\,s}^{-1}$

  • Formula / Remarks: $v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$. In one-dimensional motion, sign indicates direction.