1. Introduction: The Universe is a Movie | Motion in a Straight Line class 11 notes

Hello students! Welcome to the exciting world of Mechanics. Before we dive into formulas and graphs, let’s just sit back and think about the universe. If you look around, is anything truly “still”? The chair you are sitting on seems still, but the Earth is spinning at 1000 miles per hour. The Earth is orbiting the Sun at 67,000 miles per hour. The Sun is flying through the galaxy. So, everything is in motion.

In this chapter, we are starting with the simplest type of motion: Rectilinear Motion. This means motion along a straight line. Think of an ant walking on a tight thread, or a car driving on a perfectly straight highway. It can go forward, or it can go backward. That’s it. No turning left or right.

1.1 Kinematics vs. Dynamics

Physics is divided into branches. Right now, we are in Kinematics.

Kinematics asks: “How is it moving?” (Is it fast? Is it slowing down? Where is it?)

Dynamics (which we will study later) asks: “Why is it moving?” (Did someone push it? Is gravity pulling it?)

For this chapter, we don’t care why the car is moving. We just want to describe how it moves.

1.2 Frame of Reference: The “Who is Watching?” Problem

Imagine you are sitting in a train moving at 100 km/h. You look at your suitcase on the rack above. Is it moving? To you, it is at rest. It is not going anywhere.

Now, imagine your friend is standing on the platform outside watching the train pass by. What does he see? He sees the suitcase moving at 100 km/h!

So, who is right? Is the suitcase moving or resting?

Answer: Both are right! Motion is Relative. You cannot define motion without a Frame of Reference. A frame of reference is basically “the observer’s point of view” combined with a clock to measure time.

2. Describing Motion: Position, Path Length, and Displacement

To track an object, we need to know its location. We call this Position. We usually use the X-axis for straight-line motion. The center point (x = 0) is called the Origin.

Figure 1: Describing position on a straight line. Right of zero is positive, left is negative.

2.1 Path Length (Distance) vs. Displacement

This is the first major concept where students get confused. Let’s clear it up with a story.

The Morning Walk Story:

Imagine you wake up at Point A (Home).

1. You walk 4 km East to a park (Point B).

2. You realize you forgot your phone, so you turn back and walk 4 km West to reach Home (Point A) again.

3. Then you walk 3 km West to your school (Point C).

What is your Path Length (Distance)?

Path length is the total ground you covered. Your feet feel every step.

Calculation: 4 km (to park) + 4 km (back home) + 3 km (to school) = 11 km.

Key Feature: Path length is always positive. It never decreases. It is a Scalar quantity (has only magnitude, no direction).

What is your Displacement?

Displacement is the shortest gap between where you started and where you ended. It doesn’t care about the journey; it only cares about the result.

Start Point: Home (A).

End Point: School (C).

The gap is just 3 km to the West.

So, Displacement = -3 km (assuming West is negative).

Key Feature: Displacement is a Vector quantity (has magnitude and direction). It can be positive, negative, or even zero!

Figure 2: Distance is the full path (blue dotted line). Displacement is the shortcut (red arrow).

Real Life Example: When you swim a lap in a pool (50m there and 50m back), your distance is 100m (good exercise!), but your displacement is 0m (you are back where you started).

3. How Fast? Speed and Velocity

Just knowing “how far” isn’t enough. We need to know “how fast.” This brings us to Speed and Velocity. Just like Distance and Displacement, these two are twins with one major difference: Direction.

3.1 Average Speed

This is what your car’s odometer helps you calculate. It is simply the total distance divided by total time.

Formula: Average Speed = Total Path Length / Total Time Interval

Example: If you drive 200 km in 4 hours, your average speed is 50 km/h. It implies that, on average, you covered 50 km every hour.

3.2 Average Velocity

This tells you how fast your position is changing. It involves direction.

Formula: Average Velocity = Displacement / Total Time Interval

The Jogger’s Paradox:

A jogger runs around a circular park of radius 100m. He completes one full round in 5 minutes.

Average Speed: Total distance (Circumference 2πr) / Time. It will be a positive number.

Average Velocity: Since he came back to the start, Displacement = 0. Therefore, Average Velocity = 0 / 5 = 0 m/s.

This shows that average velocity can be zero even if the object is moving fast!

3.3 Instantaneous Velocity: The “Right Now” Speed

Average velocity is good for long trips, but it doesn’t tell the whole story. If you drive from Delhi to Agra, your average might be 60 km/h. But at some points, you might have been stuck in traffic (0 km/h) and at others speeding at 100 km/h.

The speed at a specific instant is called Instantaneous Velocity. This is what the Speedometer in your car shows.

Figure 3: The speedometer doesn’t care about your past or future. It shows your speed exactly right now.

Mathematical Definition (Calculus Alert!):

To find velocity at an exact instant, we make the time interval (t) extremely small, almost zero.

v = dx / dt

This means velocity is the derivative (rate of change) of position with respect to time. Graphically, it is the slope of the Position-Time graph at that point.

4. Acceleration: The Thrill Factor

Velocity is the rate of change of position. Acceleration is the rate of change of velocity.

Think about being in a roller coaster. When it is moving at a constant speed on a straight track, you feel normal. But when it suddenly speeds up (whoosh!) or slams on the brakes, you feel pushed against your seat. That feeling is Acceleration. Our bodies are acceleration detectors, not speed detectors.

Formula: Average Acceleration = Change in Velocity / Time Taken

a = (v – u) / t

Where ‘v’ is final velocity, ‘u’ is initial velocity, and ‘t’ is time.

Types of Acceleration

1. Positive Acceleration (Speeding Up): When you press the accelerator in a car. Velocity increases. The direction of acceleration is the SAME as the direction of motion.
2. Negative Acceleration (Retardation): When you press the brakes. Velocity decreases. The direction of acceleration is OPPOSITE to the direction of motion.
3. Zero Acceleration: This happens when velocity is constant. Cruising on a highway at a steady 80 km/h means acceleration is zero.

Figure 4: Visualizing Vectors. Blue arrow is velocity, Red arrow is acceleration. If they point the same way, you speed up. If they point opposite, you slow down.

Common Student Mistake:

“Negative acceleration always means slowing down.”

False! Imagine a car moving backwards (negative direction). If it speeds up in reverse, its velocity goes from -10 m/s to -50 m/s. This is negative acceleration, but the car is getting faster!

5. The Kinematic Equations (The “Suvat” Equations)

If an object moves with Constant Acceleration (uniform acceleration), we can predict its future perfectly. We use three famous equations derived by Newton and Galileo. These are the tools you will use to solve 90% of numericals in this chapter.

The Symbols:

u: Initial Velocity (Start speed)

v: Final Velocity (End speed)

a: Acceleration (Must be constant!)

t: Time

x (or s): Displacement

Important Note: These equations CANNOT be used if acceleration is changing (like a car driving in city traffic). In that case, you must use Calculus (Integration/Differentiation).

6. Gravity: Nature’s Constant Acceleration

The most common example of constant acceleration is gravity. If you drop a rock, the Earth pulls it. Galileo discovered that (ignoring air resistance) all objects fall with the same acceleration.

We call this acceleration due to gravity (g).

Standard value of g ≈ 9.8 m/s².

Figure 5: Free Fall. As the ball drops, gravity pulls it, making it go faster every second.

Sign Convention for Vertical Motion

To solve problems correctly, we need a rule for signs. A common convention is:

• Upward Direction: Positive (+)
• Downward Direction: Negative (-)

Since Earth always pulls DOWN, acceleration due to gravity is always:

a = -g = -9.8 m/s²

Scenario: Throwing a ball up

1. You throw it up with u = +20 m/s.

2. Gravity acts down (a = -9.8 m/s²). Since ‘u’ is positive and ‘a’ is negative, they fight. The ball slows down.

3. At the very top, the ball stops for a micro-moment. v = 0.

4. Then it falls back down. Now velocity becomes negative (downwards). Since ‘v’ is negative and ‘a’ is negative, they work together. The ball speeds up.

7. Relative Velocity: The Highway Illusion

Have you ever been on a highway? When you overtake a truck, it feels like the truck is moving very slowly backward. But if you stand on the road, the truck is moving fast forward. This is the illusion of Relative Velocity.

It measures how fast Object A is moving from the perspective of Object B.

Formula: Velocity of A with respect to B = Velocity of A – Velocity of B

v_AB = v_A – v_B

Example 1: Same Direction (Overtaking)

You (Car A) = 100 km/h. Truck (Car B) = 80 km/h. Both North.

v_AB = 100 – 80 = 20 km/h.

You see the truck moving backwards at 20 km/h relative to you.

Example 2: Opposite Direction (Collision Course)

You (Car A) = 100 km/h North. Car B = 80 km/h South (so -80 km/h).

v_AB = 100 – (-80) = 100 + 80 = 180 km/h.

This is why passing a car on a two-way road looks so scary and fast! The speeds add up.

8. Understanding Graphs: The Visual Language

Physics speaks in graphs. Being able to read a graph is a superpower. Let’s decode the three main types.

8.1 Position-Time Graph (x-t graph)

This graph tells you where the object is at any time.

• Slope of the line = Velocity. Steeper line = Faster speed.
• Horizontal Line: Object is at rest (position not changing).
• Straight Slanted Line: Uniform Motion (Constant velocity).
• Curved Line: Accelerated Motion (Speed is changing).

8.2 Velocity-Time Graph (v-t graph)

This graph tells you how fast the object is moving.

• Slope of the line = Acceleration.
• Area under the line = Displacement. This is a very powerful trick. If you calculate the area of the shape formed by the graph and the time axis, you get the distance traveled!

8.3 Acceleration-Time Graph (a-t graph)

This graph tells you the push or pull on the object.

• Area under the line = Change in Velocity.

9. Extensive Practice Set (With Teacher’s Explanations)

Physics is not a spectator sport. You have to play to understand. Here are carefully selected problems to test every concept we discussed.

Part A: Multiple Choice Questions (MCQ)

1. A car travels half the distance with speed v1 and the other half with speed v2. What is the average speed?
   (a) (v1 + v2) / 2
   (b) sqrt(v1 * v2)
   (c) 2v1v2 / (v1 + v2)
   (d) v1 / v2

   Solution: (c) 2v1v2 / (v1 + v2).
   
   Explanation: Do not just take the arithmetic mean! Average speed = Total Distance / Total Time.
   
   Let total distance be 2d.
   
   Time for first half (t1) = d/v1. Time for second half (t2) = d/v2.
   
   Total Time = d/v1 + d/v2 = d(v1+v2)/v1v2.
   
   Avg Speed = 2d / [d(v1+v2)/v1v2] = 2v1v2 / (v1+v2). This is the Harmonic Mean.

2. A stone released from an elevator going up with an acceleration ‘a’ will have what acceleration relative to the ground?
   (a) a
   (b) g
   (c) g – a
   (d) g + a

   Solution: (b) g.
   
   Explanation: Once the stone is released, it is no longer touching the elevator. No engine is pushing it. The ONLY force acting on it is Earth’s gravity. Therefore, its acceleration is simply ‘g’ downwards. The motion of the elevator doesn’t matter after release.

3. Which graph represents a ball thrown straight up and coming back down? (Velocity-Time graph)
   (a) A V-shape
   (b) A straight line with negative slope
   (c) A parabola
   (d) A horizontal line

   Solution: (b) A straight line with negative slope.
   
   Explanation: Gravity acts downwards constantly. So acceleration is constant and negative (-g). Since v = u + at, velocity depends linearly on time with a negative slope (-g). It starts positive (going up), crosses zero (top), and becomes negative (falling down).

4. The displacement of a particle is given by x = (t – 2)². What is the distance covered in the first 4 seconds?
   (a) 4 m (b) 8 m (c) 12 m (d) 0 m

   Solution: (b) 8 m.
   
   Explanation:
   
   At t=0, x = (-2)² = 4.
   
   At t=2, x = (0)² = 0. (The particle moved from 4 to 0. Distance = 4m).
   
   At t=4, x = (2)² = 4. (The particle moved from 0 to 4. Distance = 4m).
   
   Total Distance = 4 + 4 = 8m.
   
   Note: Displacement is x(4) – x(0) = 4 – 4 = 0. But distance is the actual path traveled.

Part B: Short Answer Questions

5. Is it possible for a body to have zero velocity but non-zero acceleration? Give an example.

   Answer: Yes, absolutely!
   
   Think about throwing a ball straight up. At the exact peak of its flight (maximum height), it stops for a tiny instant. Its velocity is Zero.
   
   However, gravity hasn’t stopped working! Earth is still pulling it down. So, its acceleration is 9.8 m/s² downwards. If acceleration were zero, the ball would hang in the air forever!

6. A drunkard takes 5 steps forward and 3 steps backward. Each step is 1m long and takes 1s. How long does it take him to fall in a pit 13m away?

   Answer: This is a classic pattern problem.
   
   1 cycle = 5 steps forward, 3 steps back.
   
   Net displacement per cycle = 5 – 3 = 2m.
   
   Time per cycle = 5 + 3 = 8 seconds.
   
   He needs to travel 13m.
   
   First, let’s get him close. 4 cycles = 4 * 2m = 8m. Time = 4 * 8 = 32s.
   
   Remaining distance = 13 – 8 = 5m.
   
   In the next forward stride, he walks exactly 5m.
   
   Time taken = 5 seconds.
   
   Total Time = 32s + 5s = 37 seconds.
   
   (Once he falls in the pit at 13m, he doesn’t take steps back!)

7. Why is the average velocity of a body moving in a circle with constant speed equal to zero after one complete rotation?

   Answer: Average velocity depends on Displacement. After one complete rotation, the body returns to its exact starting point.
   
   Final Position = Initial Position.
   
   Displacement = 0.
   
   Therefore, Average Velocity = 0 / Time = 0.
   
   (Average Speed, however, would be Circumference / Time).

Part C: Long Answer Questions (Numerical Solving)

8. A police jeep is chasing a thief. The jeep is moving at 45 km/h. The thief is in another car moving at 153 km/h. The police fire a bullet with a muzzle speed of 180 m/s. With what velocity will the bullet hit the thief’s car?

   Answer: This is a problem of Relative Velocity. Let’s convert everything to SI units (m/s) first.
   
   Velocity of Police (Vp) = 45 km/h = 45 * (5/18) = 12.5 m/s.
   
   Velocity of Thief (Vt) = 153 km/h = 153 * (5/18) = 42.5 m/s.
   
   Velocity of Bullet w.r.t Gun (Vb_gun) = 180 m/s.

   Step 1: Find actual velocity of the bullet.
   
   Since the gun is moving with the jeep, the bullet gets that speed too.
   
   V_bullet = Vb_gun + Vp = 180 + 12.5 = 192.5 m/s.

   Step 2: Find relative velocity of bullet w.r.t thief.
   
   This is the speed at which the bullet approaches the thief.
   
   V_hit = V_bullet – Vt
   
   V_hit = 192.5 – 42.5 = 150 m/s.
   
   So, the bullet hits the thief’s car at a speed of 150 m/s.

9. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. (Take g = 10 m/s²)

   Answer: Let’s break this into stages.

   Stage 1: First Drop
   
   u = 0, s = 90m, a = 10.
   
   Find final velocity (v): v² – u² = 2as -> v² = 2*10*90 = 1800. v = sqrt(1800) = 30sqrt(2) ≈ 42.4 m/s. (Let’s stick to simple numbers, assume v=30 m/s roughly for ease, or calculate t).
   
   Find time (t1): v = u + at -> 30sqrt(2) = 10*t1 -> t1 ≈ 4.24s.

   Stage 2: First Bounce
   
   It loses 1/10th speed. So rebound speed (u2) = 0.9 * v = 0.9 * 30sqrt(2).
   
   It goes up, stops, comes down. Time taken = 2 * (u2 / g).
   
   The graph will look like a sawtooth wave.
   
   1. From t=0 to 4.24s: Velocity goes linearly from 0 to -42.4 m/s (downward).
   
   2. At t=4.24s: Sudden jump. Velocity becomes +38.16 m/s (upward).
   
   3. Velocity decreases linearly to 0 (top of bounce) and then to negative (falling again).
   
   The slope of the line is always constant (g = 10 m/s²), except at the impact points where the line breaks vertically.

10. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

    Answer:
    
    Speed = 72 km/h = 20 m/s.
    
    Both are moving at same speed initially, so relative velocity = 0.
    
    Acceleration of B relative to A: a = 1 m/s².
    
    Time t = 50 s.
    
    We need to find the relative distance covered (s).
    
    Using s = ut + 1/2 at² (in relative frame).
    
    u_relative = 0.
    
    s = 0 * 50 + 1/2 * 1 * (50)²
    
    s = 1/2 * 2500 = 1250 m.
    
    So, the total relative distance covered is 1250 m.
    
    Since the question asks for the original distance between them (gap between Train B driver and Train A rear + lengths involved), typically “original distance between them” implies the gap.
    
    Distance = 1250 m.