Chapter 2 Notes: Motion in a Straight Line

Chapter 2: Motion in a Straight Line

2.1 Introduction

Motion is a fundamental aspect of the universe.
Examples of motion include walking, running, breathing, blood flow, falling leaves, and movement of vehicles.
The Earth rotates and revolves around the Sun; the Sun moves within the Milky Way, which itself is moving.
Motion is defined as the change in position of an object with time.
This chapter focuses on describing motion, developing the concepts of velocity and acceleration.
The study is confined to motion along a straight line (rectilinear motion).
Simple equations for rectilinear motion with uniform acceleration will be derived.
The concept of relative velocity is introduced to understand the relative nature of motion.
Objects in motion are treated as point objects, valid when the object's size is much smaller than the distance it moves.
Kinematics studies how to describe motion without delving into its causes; the causes of motion are explored later.

2.2 Instantaneous Velocity and Speed

Average velocity indicates how fast an object moves over a time interval but does not specify its speed at different instants.
Instantaneous velocity $v$ at time $t$ is defined 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 instantaneous velocity is the derivative of position with respect to time.
Graphically, instantaneous velocity at a time $t$ is the slope of the tangent to the position-time curve at that point.
Numerically, instantaneous velocity can be approximated by calculating $\frac{\Delta x}{\Delta t}$ for progressively smaller $\Delta t$ values.
Example:
- For a car's motion described by $x = 0.08t^3$ ,
- The velocity at $t = 4$ s can be found by calculating $\frac{\Delta x}{\Delta t}$ for $\Delta t$ values approaching zero (2.0 s, 1.0 s, 0.5 s, 0.1 s, 0.01 s).
- As $\Delta t$ decreases, $\frac{\Delta x}{\Delta t}$ approaches the limiting value of 3.84 m/s, which is the instantaneous velocity at $t = 4.0$ s.
It's easier to determine velocity if position data at different times or an exact expression for position as a function of time is available.
Example 2.1:
- The position of an object is given by $x = a + bt^2$ , where $a = 8.5$ m and $b = 2.5$ m/s².
- The velocity at any time $t$ is:
  - $v = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2) = 2bt = 5.0t$ m/s
- At $t = 0$ s, $v = 0$ m/s.
- At $t = 2.0$ s, $v = 10$ m/s.
- Average velocity between $t = 2.0$ s and $t = 4.0$ s:
  - $\text{Average velocity} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{(a + 16b) - (a + 4b)}{2.0} = \frac{12b}{2.0} = 6b = 6.0 \times 2.5 = 15$ m/s
For uniform motion, instantaneous velocity equals average velocity at all times.
Instantaneous speed is the magnitude of instantaneous velocity.
- A velocity of $+24.0$ m/s and a velocity of $-24.0$ m/s both have a speed of 24.0 m/s.
Average speed over a finite time interval is greater than or equal to the magnitude of average velocity.
Instantaneous speed at an instant equals the magnitude of instantaneous velocity at that instant.

2.3 Acceleration

Acceleration describes the rate of change of velocity.
Galileo concluded that the rate of change of velocity with time is constant for freely falling objects.
Average acceleration $\bar{a}$ over a time interval is the change in velocity divided by the time interval:
- $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v2 - v1}{t2 - t1}$
- $v2$ and $v1$ are instantaneous velocities at times $t2$ and $t1$ .
The SI unit of acceleration is m/s².
On a velocity-time plot, average acceleration is the slope of the line connecting points $(v2, t2)$ and $(v1, t1)$ .
Instantaneous acceleration $a$ is the limit of average acceleration as $\Delta t$ approaches zero:
- $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$
Instantaneous acceleration is the slope of the tangent to the velocity-time curve at that instant.
Acceleration can result from changes in speed, direction, or both.
Acceleration can be positive, negative, or zero.
Position-time graphs:
- Positive acceleration: curve upward.
- Negative acceleration: curve downward.
- Zero acceleration: straight line.
This chapter focuses on motion with constant acceleration.
If an object's velocity is $v_0$ at $t = 0$ and $v$ at time $t$ , then:
- $\bar{a} = \frac{v - v_0}{t - 0}$
- $v = v_0 + at$
The area under a velocity-time graph represents displacement over a time interval.
For an object moving with constant velocity $u$ , the velocity-time graph is a straight line parallel to the time axis.
- The area under the curve between $t = 0$ and $t = T$ is $u \times T = uT$ , which is the displacement.
Abrupt changes in acceleration and velocity are not physically realistic; changes are always continuous.

2.4 Kinematic Equations for Uniformly Accelerated Motion

Kinematic equations relate displacement $(x)$ , time $(t)$ , initial velocity $(v_0)$ , final velocity $(v)$ , and acceleration $(a)$ for uniformly accelerated motion.
$v = v_0 + at$
Displacement $x$ is the area under the velocity-time curve:
- $x = \frac{1}{2}(v_0 + v)t$
Substituting $v = v_0 + at$ into the displacement equation:
- $x = v_0t + \frac{1}{2}at^2$
Average velocity can be expressed as:
- $v{avg} = \frac{v0 + v}{2}$
- $x = v_{avg} t$
Substituting $t = \frac{v - v0}{a}$ into $x = \frac{(v0 + v)t}{2}$ :
- $x = \frac{(v0 + v)(v - v0)}{2a}$
- $v^2 = v_0^2 + 2ax$
The three kinematic equations are:
- $v = v_0 + at$
- $x = v_0t + \frac{1}{2}at^2$
- $v^2 = v_0^2 + 2ax$
If the initial position at $t = 0$ is $x_0$ :
- $v = v_0 + at$
- $x = x0 + v0t + \frac{1}{2}at^2$
- $v^2 = v0^2 + 2a(x - x0)$
Example 2.2:
- Derivation using Calculus:
 - Given $\frac{dv}{dt} = a$ , integrate to get $v = v_0 + at$ .
 - Given $\frac{dx}{dt} = v$ , integrate to get $x = x0 + v0t + \frac{1}{2}at^2$ .
 - Using $v \frac{dv}{dx} = a$ , integrate to get $v^2 = v0^2 + 2a(x - x0)$ .
Example 2.3:
- A ball is thrown upwards from a building top (25.0 m high) with an initial velocity of 20 m/s.
 - (a) How high will the ball rise?
 - Using $v^2 = v0^2 + 2a(y - y0)$ , where $v = 0$ , $v_0 = 20$ m/s, and $a = -10$ m/s²,
 - $0 = (20)^2 + 2(-10)(y - y_0)$
 - Solving for $(y - y_0)$ , the height above the launch point is 20 m.
 - (b) How long until the ball hits the ground?
 - Method 1: Split into upward and downward motion.
 - Upward time $t1$ : $v = v0 + at1 \implies 0 = 20 - 10t1 \implies t_1 = 2$ s.
 - Downward time $t2$ : $y = y0 + v0t + \frac{1}{2}at^2 \implies 0 = 45 + 0t2 + \frac{1}{2}(-10)t2^2 \implies t2 = 3$ s.
 - Total time $t = t1 + t2 = 2 + 3 = 5$ s.
 - Method 2: Use initial and final positions directly.
 - $y = y0 + v0t + \frac{1}{2}at^2 \implies 0 = 25 + 20t + \frac{1}{2}(-10)t^2$
 - $5t^2 - 20t - 25 = 0 \implies t = 5$ s.
Example 2.4:
- Free Fall: Motion of an object under gravity, neglecting air resistance.
  - Acceleration is constant: $a = -g = -9.8$ m/s².
  - If released from rest at $y = 0$ , then $v_0 = 0$ .
  - The equations of motion:
    - $v = -gt = -9.8t$ m/s.
    - $y = -\frac{1}{2}gt^2 = -4.9t^2$ m.
    - $v^2 = -2gy = -19.6y$ m²/s².
Example 2.5:
- Galileo’s Law of Odd Numbers: Distances traversed during equal time intervals by a body falling from rest are in the ratio 1:3:5:7…
  - Using $y = -\frac{1}{2}gt^2$ , calculate positions after equal time intervals $\tau$ .
  - The distances traveled in successive intervals $\tau$ are in the ratio 1:3:5:7:9:11…
Example 2.6:
- Stopping Distance of Vehicles: Distance a vehicle travels before stopping after applying brakes.
 - Using $v^2 = v0^2 + 2ax$ , where $v = 0$ , the stopping distance $ds$ is:
 - $ds = -\frac{v0^2}{2a}$
 - The stopping distance is proportional to the square of the initial velocity $v_0$ .
Example 2.7:
- Reaction Time: Time taken to observe, think, and act.
 - Measure reaction time by dropping a ruler and catching it.
 - Using $d = v0t + \frac{1}{2}gt^2$ , where $v0 = 0$ , the reaction time $t_r$ is:
 - $t_r = \sqrt{\frac{2d}{g}}$

Summary

Motion is the change in position with time, described with respect to an origin.
Average speed is greater than or equal to the magnitude of average velocity.
Instantaneous velocity is the limit of average velocity as $\Delta t$ approaches zero: $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$ .
Velocity at an instant is the slope of the position-time graph at that instant.
Average acceleration is the change in velocity divided by the time interval: $\bar{a} = \frac{\Delta v}{\Delta t}$ .
Instantaneous acceleration is the limit of average acceleration as $\Delta t$ approaches zero: $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}$ .
Acceleration at an instant is the slope of the velocity-time graph at that instant.
For uniform motion, acceleration is zero.
For motion with uniform acceleration, the x-t graph is a parabola, and the v-t graph is a straight line.
The area under the velocity-time curve between times $t1$ and $t2$ is equal to the displacement during that interval.
Kinematic Equations for Uniformly Accelerated Rectilinear Motion:
- $v = v_0 + at$
- $x = v_0t + \frac{1}{2}at^2$
- $v^2 = v_0^2 + 2ax$
- If the initial position is $x0$ , replace $x$ with $(x - x0)$ .

Points to Ponder

The origin and positive direction are a matter of choice; specify them before assigning signs to displacement, velocity, and acceleration.
If a particle is speeding up, acceleration is in the direction of velocity; if slowing down, acceleration is opposite to the velocity.
The sign of acceleration does not determine whether speed is increasing or decreasing; it depends on the choice of the positive direction.
Zero velocity at an instant does not necessarily imply zero acceleration at that instant.
In kinematic equations, quantities are algebraic and can be positive or negative; use proper signs.
Instantaneous velocity and acceleration definitions are always correct, while kinematic equations are true only for motion with constant magnitude and direction of acceleration.