Table of Contents
Rest and Motion
What is Rest?
A body is said to be at rest when its position does not change with respect to its surroundings (frame of reference) with the passage of time.
What is Motion?
A body is said to be in motion when its position changes with respect to its surroundings (frame of reference) with the passage of time. Rest and motion are relative terms. A body at rest in one frame of reference may be in motion in another frame of reference.
-
A body is said to be at rest if its position does not change with respect to a chosen frame of reference.
-
A body is said to be in motion if its position changes with respect to that frame of reference.
Frame of Reference
A frame of reference is a coordinate system with respect to which the position, motion, or other properties of an object are described. Since different observers can have different frames of reference, the same object can be at rest for one observer and in motion for another.
Example:
Consider a person sitting inside a moving train.
-
With respect to the train (frame of reference = train):
The person is at rest because their position relative to the seat does not change. -
With respect to the ground (frame of reference = earth):
The same person is in motion because the train (and hence the person) is moving along the track.
Point Object and Rigid Body
Point Object (Particle)
An object is considered a point object or particle when its size is negligible compared to the distance it travels. For example, Earth can be treated as a point object when studying its motion around the Sun.
Rigid Body
A rigid body is an object in which the distance between any two particles remains constant regardless of the external forces applied. Real bodies are never perfectly rigid, but many can be approximated as rigid for practical purposes.
Position and Displacement
Position: The location of a particle with respect to a chosen reference point (origin).
Displacement: The change in position of a particle. It is a vector quantity.
\[\Delta x = x_f – x_i\]
where \(x_i\) is initial position and \(x_f\) is final position.
Distance vs Displacement
Distance: Total path length covered (scalar quantity, always positive)
Displacement: Shortest distance between initial and final positions (vector quantity, can be positive, negative, or zero)
Speed and Velocity
Average Speed
\[v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}\]
Average Velocity
\[\vec{v}_{avg} = \frac{\Delta \vec{x}}{\Delta t} = \frac{x_f – x_i}{t_f – t_i}\]
Instantaneous Velocity
\[v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}\]
Acceleration
Average Acceleration
\[a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v_f – v_i}{t_f – t_i}\]
Instantaneous Acceleration
\[a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2}\]
Equations of Motion (Constant Acceleration)
For motion with uniform acceleration, we have three fundamental equations:
First Equation
\[v = u + at\]
Let:
- Initial velocity = \(u\)
- Final velocity = \(v\)
- Acceleration = \(a\)
- Time = \(t\)
Acceleration is defined as the rate of change of velocity:
\[
a = \frac{v – u}{t}
\]
Rearranging, we get:
\[
v – u = at
\]
\[
\boxed{v = u + at}
\]
This is the first equation of motion.
Second Equation
\[s = ut + \frac{1}{2}at^2\]
where \(s\) = displacement
Statement: The second equation of motion gives the relation between displacement, initial velocity, time, and acceleration.
Average velocity is given by:
\[
\text{Average velocity} = \frac{u + v}{2}
\]
Displacement is equal to average velocity multiplied by time:
\[
s = \frac{u + v}{2} \times t
\]
Using the first equation of motion \(v = u + at\), substitute for \(v\):
\[
s = \frac{u + (u + at)}{2} \times t
\]
\[
s = \frac{2u + at}{2} \times t
\]
\[
s = ut + \frac{1}{2}at^2
\]
\[
\boxed{s = ut + \frac{1}{2}at^2}
\]
This is the second equation of motion.
Third Equation
\[v^2 = u^2 + 2as\]
Statement: The third equation of motion gives the relation between final velocity, initial velocity, acceleration, and displacement.
From the first equation of motion:
\[
v = u + at
\]
Rearranging:
\[
t = \frac{v – u}{a}
\]
Substitute this value of \(t\) into the second equation of motion:
\[
s = ut + \frac{1}{2}at^2
\]
\[
s = u\left(\frac{v – u}{a}\right) + \frac{1}{2}a\left(\frac{v – u}{a}\right)^2
\]
Simplifying:
\[
2as = 2u(v – u) + (v – u)^2
\]
\[
2as = v^2 – u^2
\]
\[
\boxed{v^2 = u^2 + 2as}
\]
This is the third equation of motion.
Graphical Representation
Position-Time Graph
- Slope at any point gives instantaneous velocity: \(v = \frac{dx}{dt}\)
- Horizontal line indicates rest
- Straight line with constant slope indicates uniform motion
- Curved line indicates non-uniform motion (acceleration)
Velocity-Time Graph
- Slope at any point gives instantaneous acceleration: \(a = \frac{dv}{dt}\)
- Area under the curve gives displacement
- Horizontal line indicates uniform velocity (zero acceleration)
- Straight line indicates uniform acceleration
Acceleration-Time Graph
- Area under the curve gives change in velocity
Solved Examples
Q. A car accelerates uniformly from rest to a velocity of 30 m/s in 10 seconds. Find:
- The acceleration of the car
- The distance covered in this time
Solution:
Given: \(u = 0\) m/s, \(v = 30\) m/s, \(t = 10\) s
(a) Using \(v = u + at\)
\[30 = 0 + a(10)\]
\[a = 3 \text{ m/s}^2\]
(b) Using \(s = ut + \frac{1}{2}at^2\)
\[s = 0(10) + \frac{1}{2}(3)(10)^2\]
\[s = \frac{1}{2}(3)(100) = 150 \text{ m}\]
Q. Two trains A and B are moving on parallel tracks with velocities 72 km/h and 54 km/h respectively in the same direction. If the length of train A is 200 m and that of B is 150 m, find the time taken by A to completely overtake B.
Solution:
Converting velocities to m/s:
\[v_A = 72 \times \frac{5}{18} = 20 \text{ m/s}\]
\[v_B = 54 \times \frac{5}{18} = 15 \text{ m/s}\]
Relative velocity of A with respect to B:
\[v_{AB} = v_A – v_B = 20 – 15 = 5 \text{ m/s}\]
For complete overtaking, relative displacement needed:
\[s = L_A + L_B = 200 + 150 = 350 \text{ m}\]
Time taken:
\[t = \frac{s}{v_{AB}} = \frac{350}{5} = 70 \text{ seconds}\]
Q. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking \(g = 10\) m/s², find:
- Maximum height reached
- Time to reach maximum height
- Velocity after 3 seconds
- Total time of flight
Solution:
Taking upward direction as positive: \(u = 40\) m/s, \(a = -10\) m/s²
(a) At maximum height, \(v = 0\)
Using \(v^2 = u^2 + 2as\)
\[0 = (40)^2 + 2(-10)s\]
\[0 = 1600 – 20s\]
\[s = 80 \text{ m}\]
(b) Using \(v = u + at\)
\[0 = 40 + (-10)t\]
\[t = 4 \text{ seconds}\]
(c) Velocity after 3 seconds:
\[v = u + at = 40 + (-10)(3) = 10 \text{ m/s (upward)}\]
(d) Total time of flight = 2 × time to reach maximum height
\[T = 2 \times 4 = 8 \text{ seconds}\]
Q. A particle moves along a straight line such that its position is given by \(x = 2t^3 – 9t^2 + 12t + 3\) (in meters, t in seconds). Find:
- Velocity at t = 2 seconds
- Acceleration at t = 2 seconds
- When does the particle momentarily come to rest?
Solution:
Given: \(x = 2t^3 – 9t^2 + 12t + 3\)
(a) Velocity: \(v = \frac{dx}{dt}\)
\[v = 6t^2 – 18t + 12\]
At t = 2 s:
\[v = 6(2)^2 – 18(2) + 12 = 24 – 36 + 12 = 0 \text{ m/s}\]
(b) Acceleration: \(a = \frac{dv}{dt}\)
\[a = 12t – 18\]
At t = 2 s:
\[a = 12(2) – 18 = 6 \text{ m/s}^2\]
(c) For particle to come to rest, v = 0:
\[6t^2 – 18t + 12 = 0\]
\[t^2 – 3t + 2 = 0\]
\[(t – 1)(t – 2) = 0\]
\[t = 1 \text{ s or } t = 2 \text{ s}\]
The particle comes to rest at t = 1 s and t = 2 s.
Formula
| Quantity | Formula |
|---|---|
| Average Velocity | \(v_{avg} = \frac{\Delta x}{\Delta t}\) |
| Instantaneous Velocity | \(v = \frac{dx}{dt}\) |
| Average Acceleration | \(a_{avg} = \frac{\Delta v}{\Delta t}\) |
| Instantaneous Acceleration | \(a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\) |
| First Equation | \(v = u + at\) |
| Second Equation | \(s = ut + \frac{1}{2}at^2\) |
| Third Equation | \(v^2 = u^2 + 2as\) |
| nth Second Displacement | \(s_n = u + \frac{a}{2}(2n-1)\) |
| Average Velocity (uniform acceleration) | \(v_{avg} = \frac{u+v}{2}\) |
H C Verma Solution
1. A man has to go 50m due north, 40m due east and 20m due south to reach a field.
(a)What distance he has to walk to reach the field?
(b)What is his displacement from his house to the field?
Step 1: Identify the Path Segments
The motion consists of three distinct parts. To analyze these, we treat the movements as individual segments of a total path.
- $d_1 = 50\text{ m}$ (North)
- $d_2 = 40\text{ m}$ (East)
- $d_3 = 20\text{ m}$ (South)
Step 2: Calculate Total Distance
The sources define distance as a scalar quantity representing the total area an object covers in motion. It is calculated by summing all segments using the formula $d=d_1+d_2+d_3$.
- $d = 50\text{ m} + 40\text{ m} + 20\text{ m}$
- Total Distance = $\mathbf{110\text{ m}}$
Step 3: Establish a Coordinate System for Displacement
Displacement is a vector quantity that indicates how far an object is from its destination. According to the sources, it is expressed as $x = x_f – x_i$, where $x_f$ is the final position and $x_i$ is the initial position. We set the starting point (house) as the origin $(0,0)$.
- After moving North: $(0, 50)$
- After moving East: $(40, 50)$
- After moving South: $(40, 50 – 20) = (40, 30)$
The final position $x_f$ is $(40, 30)$.
Step 4: Calculate the Magnitude of Displacement
Using the coordinates of the initial position $(0,0)$ and final position $(40,30)$, we find the magnitude (the straight-line distance):
- $|x| = \sqrt{(40 – 0)^2 + (30 – 0)^2}$
- $|x| = \sqrt{1600 + 900} = \sqrt{2500}$
- Magnitude = $\mathbf{50\text{ m}}$
Step 5: Determine the Direction
The sources emphasize that for displacement, it is mandatory to specify the direction of travel. We find the angle $\theta$ relative to the East axis:
- $\tan \theta = \frac{\text{Vertical Displacement}}{\text{Horizontal Displacement}} = \frac{30}{40} = 0.75$
- $\theta = \tan^{-1}(0.75) \approx \mathbf{37^\circ}$
- Final Displacement: $\mathbf{50\text{ m}}$ at $\mathbf{37^\circ \text{ North of East}}$.
Q. A particle starts from the origin, goes along the X-axis to the point \( (20\,\text{m}, 0) \) and then returns along the same line to the point \( (-20\,\text{m}, 0) \). Find the distance and displacement of the particle during the trip.
Solution:
Step 1: Description of Motion
- Initial position: \( (0,0) \)
- Motion from origin to +20 m: \( 0 \rightarrow +20\,\text{m} \)
- Return motion from +20 m to −20 m: \( +20 \rightarrow -20\,\text{m} \)
(a) Distance Travelled
Distance is the total length of the actual path
travelled by the particle.
\[
\text{Distance} = 20 + 40 = \boxed{60\,\text{m}}
\]
(b) Displacement
Displacement is the change in position of the particle
from start to end.
\[
\text{Displacement} = x_f – x_i = (-20) – 0 = \boxed{-20\,\text{m}}
\]
The negative sign indicates that the displacement is along the
negative X-axis.