Motion in a Straight Line — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Straight Line" — 7 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Straight Line" — 7 important questions with detailed answers for CBSE boar…

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Key Questions Covered:

A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds, th…
A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking…
Define instantaneous velocity and average velocity. A particle moves such tha…
A train travelling at 36 km/h applies brakes and decelerates uniformly at 2 m…
What is meant by uniform acceleration? A car starting from rest reaches a vel…
Two objects A and B are moving in a straight line. Object A has initial veloc…
+ 1 more questions in the full chapter

Solutions Summary:

Question	Status
A car starts from rest and accelerates uniformly at 2 m/s…	✓ Solved
A ball is thrown vertically upward with an initial veloci…	✓ Solved
Define instantaneous velocity and average velocity. A par…	✓ Solved
A train travelling at 36 km/h applies brakes and decelera…	✓ Solved
What is meant by uniform acceleration? A car starting fro…	✓ Solved
Two objects A and B are moving in a straight line. Object…	✓ Solved

Showing 6 of 7 questions

Q1: A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds, then moves with constant velocity. Draw the v-t graph for the motion and find the total distance covered in 15 seconds.

Given: Initial velocity u = 0 m/s (starts from rest) Acceleration a = 2 m/s² Time of acceleration t₁ = 10 s Total time = 15 s Part 1: Find velocity after 10 s Using v = u + at v = 0 + 2 × 10 = 20 m/s Part 2: Distance during acceleration (0 to 10 s) Using s = ut + (1/2)at² s₁ = 0 + (1/2) × 2 × (10)² s₁ = 100 m Part 3: Motion from 10 s to 15 s Time at constant velocity t₂ = 15 - 10 = 5 s Velocity v = 20 m/s Distance s₂ = v × t₂ = 20 × 5 = 100 m Part 4: Total distance s_total = s₁ + s₂ = 10...

Q2: A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking g = 10 m/s², find (a) the time taken to reach maximum height (b) the maximum height (c) the velocity when it returns to the starting point.

Given: Initial velocity u = 20 m/s (upward, positive direction) Acceleration due to gravity g = -10 m/s² (downward, negative) Final velocity at maximum height v = 0 m/s Part (a): Time to reach maximum height Using v = u + at 0 = 20 + (-10)t 10t = 20 t = 2 s Part (b): Maximum height Using v² = u² + 2as 0² = 20² + 2(-10)h 0 = 400 - 20h 20h = 400 h = 20 m Alternative: Using s = ut + (1/2)at² h = 20(2) + (1/2)(-10)(2)² h = 40 - 20 = 20 m Part (c): Velocity when returning to starting point Tot...

Q3: Define instantaneous velocity and average velocity. A particle moves such that its position is given by x = 5t² + 3t + 2 (where x is in metres and t is in seconds). Find the instantaneous velocity at t = 2 s and average velocity from t = 0 to t = 2 s.

Definitions: Average velocity: Average velocity is the displacement divided by the total time taken. v_avg = Δx/Δt = (x₂ - x₁)/(t₂ - t₁) It represents the overall rate of change of position. Instantaneous velocity: Instantaneous velocity is the velocity at a particular instant of time. v_inst = dx/dt = lim(Δt→0) Δx/Δt It is the derivative of position with respect to time. Given: x = 5t² + 3t + 2 (where x in m, t in s) Part 1: Find instantaneous velocity at t = 2 s v = dx/dt = d/dt(5t² + 3t ...

Q4: A train travelling at 36 km/h applies brakes and decelerates uniformly at 2 m/s². How long does it take to stop and what distance does it cover before stopping?

Given: Initial velocity u = 36 km/h = 36 × (5/18) = 10 m/s Final velocity v = 0 m/s (train stops) Deceleration a = -2 m/s² (negative because it opposes motion) Part 1: Time taken to stop Using v = u + at 0 = 10 + (-2)t 2t = 10 t = 5 s Part 2: Distance covered before stopping Method 1: Using v² = u² + 2as 0² = 10² + 2(-2)s 0 = 100 - 4s 4s = 100 s = 25 m Method 2: Using s = ut + (1/2)at² s = 10(5) + (1/2)(-2)(5)² s = 50 - 25 = 25 m Method 3: Using average velocity s = [(u + v)/2] × t s = [(1...

Q5: What is meant by uniform acceleration? A car starting from rest reaches a velocity of 72 km/h in 10 seconds. Assuming uniform acceleration, find the acceleration and distance travelled.

Uniform Acceleration: Uniform acceleration (or constant acceleration) is when the velocity of a body changes at a constant rate. In other words, the acceleration remains the same throughout the motion. Mathematically: a = Δv/Δt = constant Characteristics of uniform acceleration: - Velocity changes linearly with time - v-t graph is a straight line - a-t graph is a horizontal line - The slope of the v-t graph gives the acceleration Given: Initial velocity u = 0 m/s (starts from rest) Final velo...

Q6: Two objects A and B are moving in a straight line. Object A has initial velocity 10 m/s and acceleration 2 m/s². Object B has initial velocity 20 m/s and acceleration -1 m/s². After how many seconds will both objects have the same velocity?

Given: Object A: u_A = 10 m/s a_A = 2 m/s² Object B: u_B = 20 m/s a_B = -1 m/s² Step 1: Write velocity equations for both objects For object A: v_A = u_A + a_A × t v_A = 10 + 2t For object B: v_B = u_B + a_B × t v_B = 20 + (-1)t = 20 - t Step 2: Set v_A = v_B to find time when velocities are equal 10 + 2t = 20 - t 2t + t = 20 - 10 3t = 10 t = 10/3 = 3.33 s (approximately) Step 3: Verify by calculating velocities at t = 10/3 s v_A = 10 + 2(10/3) = 10 + 20/3 = 30/3 + 20/3 = 50/3 = 16.67 m/...

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