Chapter 2 Motion in a Straight Line Notes

Chapter 2: Motion in a Straight Line

2.1 Introduction

• Motion is a fundamental aspect of the universe, exemplified by everyday actions like walking and the movement of celestial bodies.

• Motion signifies a change in an object's position over time.

• This chapter focuses on describing motion by introducing the concepts of velocity and acceleration.

• The study is limited to motion along a straight line, referred to as rectilinear motion.

• The discussion simplifies objects in motion as point objects, an approximation valid when the object's size is significantly smaller than the distance it travels.

• Kinematics is the study of describing motion without delving into its causes, which are explored in later chapters.

2.2 Instantaneous Velocity and Speed

• Average velocity indicates the rate of motion over a time interval but does not specify the velocity at particular moments.

• Instantaneous velocity, or simply velocity ($v$), is defined as the limit of the average velocity as the time interval ($\Delta t$) approaches zero:
  $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

• This is the derivative of position ($x$) with respect to time ($t$).

• Graphically, instantaneous velocity at a time $t$ is the slope of the tangent to the position-time curve at that point.

• Numerically, it can be obtained by calculating $\frac{\Delta x}{\Delta t}$ for successively smaller values of $\Delta t$.

• Example: For a car's motion described by $x = 0.08t^3$, the instantaneous velocity at $t = 4.0$ s can be found by calculating the limiting value of $\frac{\Delta x}{\Delta t}$ as $\Delta t$ approaches 0 (Table 2.1).

• Instantaneous speed is the magnitude of instantaneous velocity.

Example 2.1:

• The position of an object moving along the x-axis is given by $x = a + bt^2$, where $a = 8.5$ m and $b = 2.5$ m/s².

• To find the velocity, differentiate $x$ with respect to $t$: $v = \frac{dx}{dt} = 2bt = 5.0t$ m/s.

• At $t = 0$ s, $v = 0$ m/s, and at $t = 2.0$ s, $v = 10$ m/s.

• The average velocity between $t = 2.0$ s and $t = 4.0$ s is calculated as:
  $\text{Average velocity} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{(a + 16b) - (a + 4b)}{2.0} = \frac{12b}{2.0} = 6b = 15 \text{ m/s}$

• For uniform motion, instantaneous velocity equals average velocity.

2.3 Acceleration

• Acceleration describes the rate of change of velocity with time.

• Average acceleration ($\bar{a}$) over a time interval is defined as:
  $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v<em>2 - v</em>1}{t<em>2 - t</em>1}$

• The SI unit of acceleration is m/s².

• On a velocity-time graph, average acceleration is the slope of the line connecting two points ($(v<em>2, t</em>2)$ and $(v<em>1, t</em>1)$).

• Instantaneous acceleration ($a$) is defined as the limit of the average acceleration as the time interval 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 a given instant.

• Acceleration can be positive, negative, or zero, indicating changes in speed and/or direction.

• Position-time graphs illustrate motion with positive, negative, and zero acceleration (Figs. 2.4 a, b, c).

• This chapter primarily focuses on motion with constant acceleration.

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.

• Equation 1:
  $v = v_0 + at$

• Equation 2:
  $x = v_0t + \frac{1}{2}at^2$

• Equation 3:
  $v^2 = v_0^2 + 2ax$

• These equations assume that at $t = 0$, the position of the particle is $x = 0$. For a non-zero initial position ($x_0$), the equations are modified to:

  • $v = v_0 + at$

  • $x = x<em>0 + v</em>0t + \frac{1}{2}at^2$

  • $v^2 = v<em>0^2 + 2a(x - x</em>0)$

Example 2.2: Derivation using calculus

• Starting with $a = \frac{dv}{dt}$, integrate both sides to get $v = v_0 + at$.

• Using $v = \frac{dx}{dt}$, integrate both sides to get $x = x<em>0 + v</em>0t + \frac{1}{2}at^2$.

• Manipulate $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$, then integrate $v dv = a dx$ to derive $v^2 = v<em>0^2 + 2a(x - x</em>0)$.

Example 2.3:

• A ball is thrown upwards with an initial velocity of 20 m/s from a height of 25.0 m.

• (a) To find the maximum height, use $v^2 = v<em>0^2 + 2a(y - y</em>0)$, where $v = 0$, $v<em>0 = 20$ m/s, and $a = -10$ m/s². $0 = (20)^2 + 2(-10)(y - y</em>0)$, solving for $(y - y_0)$ gives 20 m. The ball rises 20 m above the launch point.

• (b) To find the total time, use $y = y<em>0 + v</em>0t + \frac{1}{2}at^2$, where $y<em>0 = 25$ m, $y = 0$ m, $v</em>0 = 20$ m/s, and $a = -10$ m/s².
  $0 = 25 + 20t - 5t^2$, solving the quadratic equation gives $t = 5$ s.

Example 2.4: Free Fall

• Free fall describes the motion of an object under the influence of gravity alone, neglecting air resistance.

• Acceleration due to gravity is denoted by $g$ (approximately 9.8 m/s² near the Earth's surface).

• Equations of motion for free fall (initial conditions: $y = 0$, $v_0 = 0$):

  • $v = -gt = -9.8t$ 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

• The distances traversed during equal intervals of time by a body falling from rest are in the ratio of odd numbers (1:3:5:7…).

  • This can be proven by dividing the motion into equal time intervals $\tau$ and calculating the distances traveled in each interval.

Example 2.6: Stopping Distance of Vehicles

• Stopping distance is the distance a vehicle travels after brakes are applied.

• Using $v^2 = v<em>0^2 + 2ax$, with $v = 0$, the stopping distance ($d</em>s$) is:
  $d<em>s = -\frac{v</em>0^2}{2a}$

• Stopping distance is proportional to the square of the initial velocity.

Example 2.7: Reaction Time

• Reaction time is the time it takes for someone to observe, think, and act in response to a situation.

• Reaction time can be estimated by dropping a ruler and measuring the distance it falls before being caught.

• Using $d = \frac{1}{2}gt<em>r^2$, the reaction time ($t</em>r$) is:
  $t_r = \sqrt{\frac{2d}{g}}$

Summary

• Motion is defined as the change in an object's position over time.

• Instantaneous velocity is the limit of average velocity as the time interval approaches zero: $v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$.

• Average acceleration is the change in velocity divided by the time interval: $a = \frac{\Delta v}{\Delta t}$.

• Instantaneous acceleration is the limit of average acceleration as the time interval approaches zero: $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}$.

• Area under the velocity-time curve represents the displacement.

• Kinematic equations for uniformly accelerated rectilinear motion:

  • $v = v_0 + at$

  • $x = v_0t + \frac{1}{2}at^2$

  • $v^2 = v_0^2 + 2ax$