1.0 Rest and Motion: A Relative Concept
In Physics, an object is said to be at Rest if it does not change its position with respect to its immediate surroundings. Conversely, an object is in Motion if its position changes over time relative to a reference point. Interestingly, rest and motion are relative; a passenger in a moving train is at rest relative to the carriage but in motion relative to the platform.
Scalars vs Vectors
Understanding the difference between these two is fundamental to Kinematics:
- Scalar Quantities: Have only magnitude (e.g., Distance, Speed, Mass).
- Vector Quantities: Have both magnitude and direction (e.g., Displacement, Velocity, Acceleration).
1.1 Distance and Displacement
While they may sound similar, they represent very different physical concepts in one-dimensional motion:
| Feature | Distance | Displacement |
|---|---|---|
| Definition | Total path length traveled by the body. | Shortest straight-line distance between start and end. |
| Type | Scalar | Vector |
| Value | Always positive. | Can be positive, negative, or zero. |
Speed and Velocity
$$Average\,Speed = \frac{\text{Total Distance}}{\text{Total Time}}$$
$$\vec{v} = \frac{\vec{s}}{t}$$
Where: $\vec{v}$ = Velocity, $\vec{s}$ = Displacement, $t$ = Time.
If a body travels in a circular path and returns to its starting point, the distance is $2\pi r$, but the displacement is exactly zero. Always look for the final position relative to the initial position!
A boy walks 3 km North and then 4 km East. Calculate his total distance and total displacement.
Solution:
1. Total Distance: $3\,km + 4\,km = 7\,km$.
2. Displacement: Using Pythagoras theorem for the right-angled triangle formed:
$s = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,km$.
Final Answer: Distance = $7\,km$; Displacement = $5\,km$ (towards North-East).
The Odometer in your car measures the total distance traveled, while the Speedometer measures instantaneous speed. Neither of them provides information about displacement or direction!
2.0 Acceleration: Change in Velocity
In the real world, objects rarely move at a constant velocity. When you press the gas pedal of a car, it speeds up; when you hit the brakes, it slows down. This change in velocity over time is what we call Acceleration.
Uniform vs. Non-Uniform Acceleration
- Uniform Acceleration: Velocity changes by equal amounts in equal intervals of time (e.g., a ball falling freely under gravity).
- Non-Uniform Acceleration: Velocity changes by unequal amounts in equal intervals of time (e.g., a car moving in heavy traffic).
- Retardation (Deceleration): If the velocity of a body decreases with time, the acceleration is negative.
Formula for Acceleration
$$a = \frac{v - u}{t}$$
Where: $a$ = acceleration, $v$ = final velocity, $u$ = initial velocity, $t$ = time taken.
S.I. Unit: $m/s^2$
2.1 Graphical Representation of Motion
Graphs are a powerful tool to visualize motion. For ICSE Class 9, we focus on two main types:
| Graph Type | Slope Represents | Area Represents |
|---|---|---|
| Displacement-Time ($s-t$) | Velocity | N/A |
| Velocity-Time ($v-t$) | Acceleration | Displacement |
If a body is slowing down, your calculation for '$a$' will result in a negative number. When writing the answer, you can say "Acceleration is $-2\,m/s^2$" OR "Retardation is $2\,m/s^2$". Never use "Negative Retardation" as it is a double negative!
A car starts from rest and acquires a velocity of $54\,km/h$ in $10$ seconds. Calculate its acceleration.
Solution:
1. Initial Velocity ($u$): $0\,m/s$ (Starts from rest).
2. Final Velocity ($v$): $54\,km/h = 54 \times \frac{5}{18} = 15\,m/s$.
3. Time ($t$): $10\,s$.
4. Calculation: $a = \frac{v - u}{t} = \frac{15 - 0}{10} = 1.5\,m/s^2$.
Final Answer: The acceleration of the car is $1.5\,m/s^2$.
Galileo Galilei was the first to realize that in a vacuum, all objects fall with the same constant acceleration ($g \approx 9.8\,m/s^2$), regardless of their mass. He proved this by rolling balls down inclined planes!
3.0 Equations of Motion
For a body moving with uniform acceleration in a straight line, there are three fundamental equations that relate displacement ($s$), initial velocity ($u$), final velocity ($v$), acceleration ($a$), and time ($t$). These are known as the Kinematic Equations.
The Three Equations of Motion
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$
Note: These equations are valid only when acceleration ($a$) is constant.
3.1 Motion Under Gravity
When a body falls freely towards the Earth, it experiences a uniform acceleration called Acceleration due to Gravity ($g$). For Earth, $g \approx 9.8\,m/s^2$ (often taken as $10\,m/s^2$ for numerical ease).
Sign Conventions for Gravity
- Body Falling Downwards: The velocity increases, so $a = +g$.
- Body Thrown Upwards: The velocity decreases, so $a = -g$.
- At Maximum Height: The final velocity ($v$) of the body is always zero.
1. "Starts from rest" $\rightarrow u = 0$
2. "Comes to a stop" or "Brakes are applied" $\rightarrow v = 0$
3. "Moves with uniform velocity" $\rightarrow a = 0$
A stone is dropped from a cliff of height $20\,m$. Calculate the time it takes to reach the ground and its velocity on hitting the ground. (Take $g = 10\,m/s^2$)
Solution:
1. Given: $u = 0$ (dropped), $s = 20\,m$, $a = g = 10\,m/s^2$.
2. Finding Time ($t$): Use $s = ut + \frac{1}{2}at^2$
$20 = 0 + \frac{1}{2}(10)t^2 \Rightarrow 20 = 5t^2 \Rightarrow t^2 = 4 \Rightarrow \mathbf{t = 2\,s}$.
3. Finding Final Velocity ($v$): Use $v = u + at$
$v = 0 + (10)(2) \Rightarrow \mathbf{v = 20\,m/s}$.
Final Answer: Time = $2$ seconds; Velocity = $20\,m/s$.
If you throw a ball upwards with a certain speed, it will return to your hand with the exact same speed (ignoring air resistance). The time taken to go up is also exactly equal to the time taken to come down!
4.0 Graphical Derivation of Equations of Motion
In ICSE Physics, deriving the equations of motion using a Velocity-Time ($v-t$) graph is a vital skill. We consider a body moving with an initial velocity $u$ at time $t=0$, which accelerates uniformly to reach a final velocity $v$ in time $t$.
Graph Interpretation
- The Slope: The slope of the line in a $v-t$ graph represents the Acceleration ($a$).
- The Area: The area under the slope line and the time axis represents the Displacement ($s$).
4.1 The Step-by-Step Derivations
1. To derive $v = u + at$:
Acceleration = Slope of the graph = $\frac{\text{Change in Velocity}}{\text{Time}}$
$a = \frac{v - u}{t} \Rightarrow at = v - u \Rightarrow \mathbf{v = u + at}$
2. To derive $s = ut + \frac{1}{2}at^2$:
Displacement ($s$) = Area of Trapezoid = Area of Rectangle ($u \times t$) + Area of Triangle ($\frac{1}{2} \times t \times (v-u)$).
Substitute $(v-u) = at$ from the first equation:
$s = ut + \frac{1}{2}(t)(at) \Rightarrow \mathbf{s = ut + \frac{1}{2}at^2}$
Alternative Displacement Formula
Displacement is also the product of average velocity and time:
$$s = \left(\frac{u + v}{2}\right) \times t$$
When a body moves in a circle with constant speed, its velocity is still changing because its direction changes at every point. Therefore, uniform circular motion is always an accelerated motion, even if the speed is constant!
From a $v-t$ graph, the area of a triangle is $50\,m$ and the area of the rectangle below it is $150\,m$. What is the total displacement?
Solution:
1. Total Displacement ($s$) = Total area under the $v-t$ curve.
2. $s = \text{Area of Rectangle} + \text{Area of Triangle}$.
3. $s = 150\,m + 50\,m = 200\,m$.
Final Answer: The total displacement is $200\,m$.
Modern GPS systems calculate your position using complex versions of these equations! By measuring the time it takes for signals to travel from satellites to your phone, the device calculates your displacement from multiple points to find your exact coordinates.