Kinematics: Motion in a Straight Line

Welcome to Kinematics! This is the branch of physics where we study how things move, without worrying about why they move (we don't care about forces here; that comes in the next chapter). Let's start from the absolute basics and build up to Board Exam levels!

1. The Concept of Rest and Motion

In physics, nothing is in absolute rest or absolute motion. It all depends on who is looking! This is called the Frame of Reference.

Concept Rest: An object is at rest if its position does not change with time relative to an observer.

Motion: An object is in motion if its position changes with time relative to an observer.

Fact Example: If you are sitting inside a moving train, you are at rest with respect to your co-passenger, but you are in motion with respect to a person standing outside on the platform!

Practice Problems: Rest & Motion

A person is sitting in a moving car. Is he at rest or in motion with respect to (a) the car seat, (b) a tree outside the window?
(Ans: (a) Rest, (b) Motion)
Can an object be in motion and at rest simultaneously? Explain.
(Hint: Yes, relative to different frames of reference. Give the train example.)
Two trains A and B are moving in the same direction with the exact same speed. What is the state of train A with respect to a passenger sitting in train B?
(Ans: At rest)

2. Distance vs. Displacement

This is the most fundamental difference in kinematics. Students often confuse the two, so pay close attention!

Distance

It is the actual length of the path traveled by an object.
It is a Scalar Quantity (it only has magnitude, no direction. Example: 50 km).
It can never be negative or zero for a moving object.

Displacement

It is the shortest straight-line distance between the initial and final positions.
It is a Vector Quantity (it has magnitude AND direction. Example: 50 km North).
It can be positive, negative, or zero!

Important Golden Rule for Board Exams:
Distance is always greater than or equal to the magnitude of Displacement.
\( \text{Distance} \ge |\text{Displacement}| \)
They are only equal when the object moves in a perfectly straight line without turning back!

Difference Between Distance and Displacement

Practice Problems: Distance & Displacement

A student walks 3 km East, then turns and walks 4 km North. Find the total distance traveled and the displacement.
(Ans: Distance = 7 km, Displacement = 5 km using Pythagoras theorem)
An athlete completes one round of a circular track of radius \( R \) in 40 seconds. What will be his displacement at the end of 2 minutes 20 seconds? (NCERT)
(Hint: 2 min 20 sec = 140s. Number of rounds = 140/40 = 3.5. He ends up exactly opposite to the starting point. Ans: \( 2R \))
Can displacement be zero even if distance is not zero? Give a real-life example.
(Ans: Yes, like walking to school and returning back home.)
A boy throws a ball straight up to a height of 10 m, and it falls back into his hands. What is the distance and displacement of the ball?
(Ans: Distance = 20 m, Displacement = 0 m)

3. Speed and Velocity

Just like distance and displacement, we have speed and velocity to tell us how fast an object is moving.

Concept Speed: The rate of change of distance. (Scalar)
\( v = \frac{\text{Distance}}{\text{Time}} \)

Velocity: The rate of change of displacement. (Vector)
\( \vec{v} = \frac{\text{Displacement}}{\text{Time}} \)

Types of Speed/Velocity

Uniform: Traveling equal distances/displacements in equal intervals of time.
Average: Total distance divided by total time.
Instantaneous: Speed or velocity at a specific, exact moment in time (like what a car's speedometer shows). Mathematically, it's the derivative of position: \( v = \frac{dx}{dt} \).

Practice Problems: Speed & Velocity

Convert the speed of a train traveling at 72 km/h into m/s.
(Hint: Multiply by 5/18. Ans: 20 m/s)
A car travels from City A to City B at a speed of 40 km/h and returns from B to A at 60 km/h. Calculate the average speed and average velocity of the car. (NCERT)
(Hint: Average velocity is 0 because displacement is 0. For average speed use \( \frac{2v_1v_2}{v_1+v_2} \). Ans: 48 km/h)
Odometer of a car reads 2000 km at the start of a trip and 2400 km at the end. If the trip took 8 hours, calculate average speed in km/h and m/s.
(Ans: 50 km/h, 13.9 m/s)

4. Acceleration

When you press the gas pedal on a car, your velocity changes. This change in velocity over time is called Acceleration.

Concept Acceleration (\( a \)): The rate of change of velocity with respect to time. It is a vector quantity.
\( a = \frac{v - u}{t} \)
Where \( v \) is final velocity, \( u \) is initial velocity, and \( t \) is time taken.

SI Unit: \( m/s^2 \)

Important What is Retardation?
If velocity is decreasing, the acceleration acts in the opposite direction to the motion. This is called negative acceleration, deceleration, or retardation. (Example: applying brakes).

Practice Problems: Acceleration

A train starting from a railway station (initially at rest) moves with uniform acceleration and attains a speed of 40 km/h in 10 minutes. Find its acceleration in \( m/s^2 \).
(Hint: Convert km/h to m/s, and minutes to seconds first. \( u=0 \). Ans: \( \approx 0.0185 \text{ } m/s^2 \))
A motorcycle moving with a speed of 5 m/s is subjected to an acceleration of \( 0.2 \text{ } m/s^2 \). Calculate the speed of the motorcycle after 10 seconds.
(Hint: Use \( v = u + at \). Ans: 7 m/s)
Is it possible for an object to have zero velocity but non-zero acceleration?
(Hint: Yes, think of a ball thrown straight up at its absolute highest point. Velocity is 0, but gravity \( 9.8 \text{ } m/s^2 \) is still acting on it!)

5. Kinematic Equations for Uniform Acceleration

These three equations are the heart of Class 11 mechanics. You will use them everywhere! Note: They can ONLY be used when acceleration (\( a \)) is constant.

First Equation (Velocity-Time Relation):
\( v = u + at \)
Second Equation (Position-Time Relation):
\( s = ut + \frac{1}{2}at^2 \)
Third Equation (Velocity-Position Relation):
\( v^2 = u^2 + 2as \)

Where: \( u \) = initial velocity, \( v \) = final velocity, \( a \) = acceleration, \( t \) = time, \( s \) = displacement.

Practice Problems: Kinematic Equations

A racing car has a uniform acceleration of \( 4 \text{ } m/s^2 \). What distance will it cover in 10 seconds after the start?
(Hint: Starts from rest means \( u = 0 \). Use 2nd equation. Ans: 200 m)
A scooter moving at a speed of 10 m/s is stopped by applying brakes which produce a uniform retardation of \( -0.5 \text{ } m/s^2 \). How much distance will be covered by the scooter before it stops?
(Hint: Stops means \( v = 0 \). Use 3rd equation. Ans: 100 m)
An electron moving with a velocity of \( 5 \times 10^4 \text{ } m/s \) enters a uniform electric field and acquires a uniform acceleration of \( 10^4 \text{ } m/s^2 \). Calculate the time in which its velocity doubles.
(Hint: Final velocity \( v \) will be \( 10 \times 10^4 \text{ } m/s \). Use 1st equation. Ans: 5 seconds)

6. Graphical Representation of Motion

Graphs are a visual way to understand kinematics. Let's look at the two most important ones.

A. Position-Time (x-t) Graph

The slope of a position-time graph gives the Velocity.
Steeper slope = Higher velocity.
Horizontal line = Object is at rest (zero velocity).

B. Velocity-Time (v-t) Graph

The slope of a velocity-time graph gives the Acceleration.
The Area under the curve of a velocity-time graph gives the Displacement. This is extremely important for board exams!

Fact If the v-t graph is a straight line sloping upwards, it means the object has uniform positive acceleration.

Practice Problems: Graphs

What does the area under a velocity-time graph represent?
(Ans: Displacement)
What does the slope of a position-time graph represent?
(Ans: Velocity)
If the position-time graph of a moving object is a straight horizontal line parallel to the time axis, what is the object's state?
(Ans: The object is at rest, velocity is zero)
Draw (on paper) a velocity-time graph for a car moving with a constant speed of 20 m/s for 5 seconds, and then applying brakes to stop uniformly in the next 5 seconds. Calculate total displacement using the area.
(Hint: Area of rectangle + Area of triangle. Ans: 100m + 50m = 150m)

7. Motion Under Gravity (Free Fall)

When an object is dropped, thrown up, or thrown down, the earth pulls it with a constant acceleration called acceleration due to gravity (\( g \)).

Value of \( g \approx 9.8 \text{ } m/s^2 \) (often taken as \( 10 \text{ } m/s^2 \) for easy calculations).

Important Sign Convention Trick for Problem Solving:
Always choose a direction as positive.
Easiest method: Take the upward direction as Positive (+) and downward as Negative (-).
Therefore, gravity (\( g \)) is ALWAYS pointing down, so \( a = -g \) in your equations!

When an object is thrown vertically upwards:

At the highest point, its velocity becomes zero (\( v = 0 \)).
Time of ascent = Time of descent (if air resistance is ignored).
The kinematic equations become:
\( v = u - gt \)
\( s = ut - \frac{1}{2}gt^2 \)
\( v^2 = u^2 - 2gs \)

Physics Diagram: Object Thrown Vertically Upward

Practice Problems: Motion Under Gravity

A ball is thrown vertically upwards with a velocity of 20 m/s. How high will it go? (Take \( g = 10 \text{ } m/s^2 \)).
(Hint: At max height \( v = 0 \). Acceleration \( a = -10 \). Use \( v^2 = u^2 + 2as \). Ans: 20 m)
A stone is dropped from a cliff of height 20 m. How much time will it take to reach the ground? (Take \( g = 10 \text{ } m/s^2 \)).
(Hint: Dropped means \( u = 0 \). Displacement is downward so \( s = -20 \). \( a = -10 \). Use 2nd equation. Ans: 2 seconds)
A ball thrown up is caught by the thrower after 4 seconds. With what velocity was it thrown?
(Hint: Total time is 4s, so time to reach max height is 2s. At max height \( v=0, t=2 \). Use 1st equation \( v = u - gt \). Ans: \( u = 20 \text{ } m/s \))