Chapter 2 Motion in a Straight Line Notes

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.

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.

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{v2 - v1}{t2 - t1}
The SI unit of acceleration is m/s².
On a velocity-time graph, average acceleration is the slope of the line connecting two points ((v2, t2) and (v1, t1)).
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.

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 = x0 + v0t + \frac{1}{2}at^2
- v^2 = v0^2 + 2a(x - x0)

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 = x0 + v0t + \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 = v0^2 + 2a(x - x0).

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 = v0^2 + 2a(y - y0), where v = 0, v0 = 20 m/s, and a = -10 m/s². 0 = (20)^2 + 2(-10)(y - y0), 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 = y0 + v0t + \frac{1}{2}at^2, where y0 = 25 m, y = 0 m, v0 = 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 = v0^2 + 2ax, with v = 0, the stopping distance (ds) is:
ds = -\frac{v0^2}{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}gtr^2, the reaction time (tr) is:
t_r = \sqrt{\frac{2d}{g}}

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