NCERT Class 11 Physics Chapter 3 Exercise Solutions - Motion in a Straight Line

NCERT Class 11 Physics Chapter 3 - Exercise Solutions

Motion in a Straight Line

Welcome to comprehensive exercise solutions for NCERT Class 11 Physics Chapter 3: Motion in a Straight Line. These solutions provide step-by-step explanations for all textbook exercises, helping you master the concepts of kinematics.

📚 Exercise 3.1 - Multiple Choice Questions

Question 1

Among the following, which is not a vector quantity?

(a) Displacement
(b) Velocity
(c) Acceleration
(d) Distance

Solution: Answer: (d) Distance

Explanation:

Vector quantities have both magnitude and direction
Scalar quantities have only magnitude

Analysis of options:

Displacement: Has magnitude and direction → Vector ✗
Velocity: Has magnitude and direction → Vector ✗
Acceleration: Has magnitude and direction → Vector ✗
Distance: Has only magnitude, no specific direction → Scalar ✓

Key Concept: Distance is the total path length traveled, while displacement is the change in position from initial to final point.

Question 2

Which of the following statements is incorrect?

(a) Displacement can be zero while distance is non-zero
(b) Displacement can be positive while distance is always positive
(c) Distance is always greater than or equal to displacement
(d) Distance and displacement are always equal

Solution: Answer: (d) Distance and displacement are always equal

Explanation: Let’s analyze each statement:

(a) Displacement can be zero while distance is non-zero

Example: A person walks 10m forward and 10m backward
Displacement: Final position - Initial position = 0m
Distance: Total path traveled = 10m + 10m = 20m
This statement is correct ✗

(b) Displacement can be positive while distance is always positive

Displacement: Can be positive, negative, or zero depending on direction
Distance: Always positive (or zero) as it’s a measure of path length
This statement is correct ✗

(c) Distance is always greater than or equal to displacement

Mathematically: Distance ≥ |Displacement|
Example: In straight line motion in one direction, distance = displacement
In any other case: distance > displacement
This statement is correct ✗

(d) Distance and displacement are always equal

This is incorrect ✓
Counterexample: Circular motion of 360°
- Distance = circumference = 2πr
- Displacement = 0 (returns to starting point)

Question 3

A particle starts from rest and moves with uniform acceleration. The ratio of distances covered in the first, second, and third seconds is:

(a) 1:2:3
(b) 1:3:5
(c) 1:4:9
(d) 1:1:1

Solution: Answer: (b) 1:3:5

Explanation: For uniformly accelerated motion starting from rest:

Formula: Distance covered in nth second = u + a(n - 0.5)

Given:

Initial velocity, u = 0 (starts from rest)
Uniform acceleration, a = constant

Distance in 1st second (n=1): s₁ = 0 + a(1 - 0.5) = 0.5a

Distance in 2nd second (n=2): s₂ = 0 + a(2 - 0.5) = 1.5a

Distance in 3rd second (n=3): s₃ = 0 + a(3 - 0.5) = 2.5a

Ratio: s₁ : s₂ : s₃ = 0.5a : 1.5a : 2.5a = 1 : 3 : 5

Alternative Method Using Kinematic Equations:

Distance traveled in time t: s = ut + ½at²

Distance in first 1 second: s₁ = 0 + ½a(1)² = 0.5a Distance in first 2 seconds: s₂ = 0 + ½a(2)² = 2a Distance in first 3 seconds: s₃ = 0 + ½a(3)² = 4.5a

Distance in individual seconds:

1st second: s₁ = 0.5a
2nd second: s₂ - s₁ = 2a - 0.5a = 1.5a
3rd second: s₃ - s₂ = 4.5a - 2a = 2.5a

Ratio: 0.5a : 1.5a : 2.5a = 1 : 3 : 5 ✓

📚 Exercise 3.2 - Very Short Answer Questions

Question 1

What is the difference between distance and displacement?

Solution:

Distance:

Definition: Total path length traveled by an object
Nature: Scalar quantity (has magnitude only)
Value: Always positive or zero
Example: If you walk 5m east and 3m west, distance = 5 + 3 = 8m

Displacement:

Definition: Change in position from initial to final point
Nature: Vector quantity (has magnitude and direction)
Value: Can be positive, negative, or zero
Example: If you walk 5m east and 3m west, displacement = 5 - 3 = 2m east

Key Difference: Distance measures how much ground was covered, while displacement measures how far and in what direction the object moved from its starting point.

Question 2

Can the magnitude of displacement be greater than the distance travelled by an object?

Solution:

No, the magnitude of displacement can never be greater than the distance travelled.

Mathematical Relationship: |Displacement| ≤ Distance

Reasoning:

Displacement is the shortest distance between initial and final positions
Distance is the actual path length traveled
The shortest path between two points is always less than or equal to any other path

Examples:

Straight line motion: Distance = Displacement (minimum distance achieved)
Curved path motion: Distance > Displacement
Return to starting point: Distance > 0, Displacement = 0

Conclusion: Distance is always greater than or equal to the magnitude of displacement.

📚 Exercise 3.3 - Short Answer Questions

Question 1

A car travels from point A to point B with a speed of 40 km/h and returns from B to A with a speed of 60 km/h. Find the average speed for the entire journey.

Solution:

Given:

Speed from A to B = 40 km/h
Speed from B to A = 60 km/h
Distance between A and B = d (let’s assume)

Step 1: Calculate time taken for each part of the journey

Time from A to B: t₁ = Distance/Speed = d/40 hours

Time from B to A: t₂ = Distance/Speed = d/60 hours

Step 2: Calculate total distance and total time

Total distance: Total Distance = d + d = 2d

Total time: Total Time = t₁ + t₂ = d/40 + d/60

Step 3: Find common denominator and add Total Time = (3d + 2d)/120 = 5d/120 = d/24 hours

Step 4: Calculate average speed Average Speed = Total Distance/Total Time Average Speed = 2d ÷ (d/24) Average Speed = 2d × (24/d) Average Speed = 48 km/h

Answer: The average speed for the entire journey is 48 km/h.

Verification: The average speed (48 km/h) is indeed between the two speeds (40 km/h and 60 km/h), which makes physical sense.

Question 2

A ball is thrown vertically upward with a velocity of 20 m/s. Find the maximum height reached and the time taken to reach the maximum height. (Take g = 10 m/s²)

Solution:

Given:

Initial velocity, u = 20 m/s (upward)
Final velocity at maximum height, v = 0 m/s
Acceleration due to gravity, g = 10 m/s² (downward, so a = -10 m/s²)
We need to find: Maximum height (s) and Time to reach maximum height (t)

Step 1: Use equation v² = u² + 2as to find maximum height

At maximum height:

Final velocity, v = 0
Initial velocity, u = 20 m/s
Acceleration, a = -10 m/s²

Substitute in equation: 0² = 20² + 2(-10)s 0 = 400 - 20s 20s = 400 s = 400/20 = 20 m

Maximum height reached = 20 meters

Step 2: Use equation v = u + at to find time taken

Substitute in equation: 0 = 20 + (-10)t 0 = 20 - 10t 10t = 20 t = 20/10 = 2 s

Time taken to reach maximum height = 2 seconds

Answer:

Maximum height reached = 20 m
Time taken to reach maximum height = 2 s

Verification: Time to go up = 2s, Time to come down = 2s (due to symmetry) Total time of flight = 4s, which is reasonable for the given initial velocity.

📚 Exercise 3.4 - Long Answer Questions

Question 1

Derive the equations of motion for uniformly accelerated motion using graphical method.

Solution:

Uniformly Accelerated Motion: Motion with constant acceleration

Given:

Initial velocity = u
Final velocity = v
Acceleration = a (constant)
Time = t
Displacement = s

Graphical Derivation:

Step 1: Velocity-Time Graph

Plot velocity (v) on y-axis and time (t) on x-axis
The graph is a straight line with slope = a (acceleration)
The area under the graph gives displacement

Step 2: Derive First Equation (v = u + at)

From the graph:

Slope = (v - u)/(t - 0) = a
Therefore: a = (v - u)/t
Rearranging: v - u = at
Final equation: v = u + at

Step 3: Derive Second Equation (s = ut + ½at²)

Area under velocity-time graph = displacement

The area consists of:
- Rectangle: base = t, height = u → Area = ut
- Triangle: base = t, height = v-u = at → Area = ½ × t × at = ½at²

Total displacement: s = ut + ½at²

Step 4: Derive Third Equation (v² = u² + 2as)

From v = u + at: t = (v - u)/a

Substitute t in s = ut + ½at²: s = u[(v - u)/a] + ½a[(v - u)/a]² s = (uv - u²)/a + ½(v - u)²/a as = uv - u² + ½(v² - 2uv + u²) as = uv - u² + ½v² - uv + ½u² as = -u² + ½v² + ½u² as = -½u² + ½v² 2as = -u² + v² v² = u² + 2as

Final Equations of Motion:

v = u + at
s = ut + ½at²
v² = u² + 2as

These equations are valid only for uniformly accelerated motion.

🎯 Key Takeaways

Important Formulas:

Average Speed: Total Distance/Total Time
First Equation: v = u + at
Second Equation: s = ut + ½at²
Third Equation: v² = u² + 2as
Distance in nth second: sₙ = u + a(n - 0.5)

Problem-Solving Strategy:

Identify given quantities
Choose appropriate equation
Pay attention to sign conventions
Check if answer is physically reasonable

Common Mistakes to Avoid:

Sign Convention: Always be consistent with positive/negative directions
Units: Ensure all quantities are in consistent units
Context: Understand whether you’re dealing with distance or displacement
Applicability: Use equations only for uniformly accelerated motion

Remember: Practice is key to mastering kinematics. Solve as many problems as possible to build confidence and speed.

Happy Learning! 🚀