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.
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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NCERT Solutions for Class 11 Physics — Motion in a Straight Line. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Motion in a Straight Line — All 7 Questions Solved
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 = 100 + 100 = 200 m
Part 5: v-t graph description
The graph has two parts:
- From t = 0 to t = 10 s: straight line from (0,0) to (10,20) with slope = 2 m/s²
- From t = 10 to t = 15 s: horizontal line at v = 20 m/s
Answer: Total distance covered = 200 m
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
Total time to return = 2 × 2 = 4 s
Using v = u + at
v = 20 + (-10)(4)
v = 20 - 40 = -20 m/s
The negative sign indicates downward direction.
Speed = 20 m/s (downward)
Answers:
(a) Time to reach maximum height = 2 s
(b) Maximum height = 20 m
(c) Velocity when returning = 20 m/s downward (or -20 m/s if upward is positive)
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 + 2)
v = 10t + 3
At t = 2 s:
v = 10(2) + 3 = 20 + 3 = 23 m/s
Part 2: Find average velocity from t = 0 to t = 2 s
Position at t = 0:
x₀ = 5(0)² + 3(0) + 2 = 2 m
Position at t = 2:
x₂ = 5(2)² + 3(2) + 2 = 5(4) + 6 + 2 = 20 + 6 + 2 = 28 m
Displacement Δx = x₂ - x₀ = 28 - 2 = 26 m
Time interval Δt = 2 - 0 = 2 s
Average velocity = 26/2 = 13 m/s
Answers:
Instantaneous velocity at t = 2 s = 23 m/s
Average velocity from t = 0 to t = 2 s = 13 m/s
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 = [(10 + 0)/2] × 5 = 5 × 5 = 25 m
Answers:
Time taken to stop = 5 s
Distance covered = 25 m
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 velocity v = 72 km/h = 72 × (5/18) = 20 m/s
Time t = 10 s
Part 1: Calculate acceleration
Using v = u + at
20 = 0 + a(10)
a = 20/10 = 2 m/s²
Part 2: Calculate distance travelled
Method 1: Using s = ut + (1/2)at²
s = 0 + (1/2)(2)(10)²
s = 1 × 100 = 100 m
Method 2: Using v² = u² + 2as
(20)² = 0² + 2(2)s
400 = 4s
s = 100 m
Method 3: Using average velocity
s = [(u + v)/2] × t = [(0 + 20)/2] × 10 = 10 × 10 = 100 m
Answers:
Acceleration = 2 m/s²
Distance travelled = 100 m
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/s
v_B = 20 - (10/3) = 60/3 - 10/3 = 50/3 = 16.67 m/s
Both velocities are equal, confirming our answer.
Answer: After 10/3 seconds or 3.33 seconds (or 3⅓ seconds), both objects will have the same velocity of 50/3 m/s or 16.67 m/s
Q7. A particle moves with position given by x = 2t³ - 3t² + 4t + 1 (x in m, t in s). Find (a) velocity at t = 1 s (b) acceleration at t = 1 s (c) describe the motion at t = 1 s.
Given:
x = 2t³ - 3t² + 4t + 1
Part (a): Find velocity at t = 1 s
Velocity v = dx/dt
v = d/dt(2t³ - 3t² + 4t + 1)
v = 6t² - 6t + 4
At t = 1 s:
v = 6(1)² - 6(1) + 4
v = 6 - 6 + 4 = 4 m/s
The velocity is positive, so the particle is moving in the positive direction.
Part (b): Find acceleration at t = 1 s
Acceleration a = dv/dt = d²x/dt²
a = d/dt(6t² - 6t + 4)
a = 12t - 6
At t = 1 s:
a = 12(1) - 6
a = 12 - 6 = 6 m/s²
The acceleration is positive, so it is in the positive direction.
Part (c): Describe the motion at t = 1 s
At t = 1 s:
- Velocity v = 4 m/s (positive) → particle moving in positive direction
- Acceleration a = 6 m/s² (positive) → acceleration in same direction as velocity
Conclusion: At t = 1 s, the particle is moving in the positive direction with speed 4 m/s, and is accelerating in the same direction. Therefore, the speed is increasing. The motion is accelerated motion in the positive direction.
Answers:
(a) Velocity at t = 1 s = 4 m/s
(b) Acceleration at t = 1 s = 6 m/s²
(c) The particle is accelerating in the direction of motion, so its speed is increasing
Motion in a Straight Line — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A particle moves along a straight line. Its position at time t is given by x = 2t + 3. What is its velocity?
- A. 2 m/s
- B. 3 m/s
- C. (2t+3) m/s
- D. 0 m/s
Answer: A. 2 m/s — The position equation is x = vt + x₀. Comparing with x = 2t + 3, the velocity v is 2 m/s. Alternatively, velocity is the derivative of position with respect to time, d(2t+3)/dt = 2 m/s.
MCQ 2. A car accelerates uniformly from rest at 2 m/s² for 10 seconds. What is the distance covered by the car during this time?
- A. 20 m
- B. 100 m
- C. 200 m
- D. 10 m
Answer: B. 100 m — Using the kinematic equation s = ut + (1/2)at², with u=0, a=2 m/s², t=10 s, we get s = 0*(10) + (1/2)*2*(10)² = 100 m.
MCQ 3. Which of the following physical quantities is a scalar?
- A. Velocity
- B. Displacement
- C. Distance
- D. Acceleration
Answer: C. Distance — Distance is the total path length covered and only has magnitude, making it a scalar quantity. Velocity, displacement, and acceleration all have both magnitude and direction, making them vector quantities.
MCQ 4. A stone is dropped from the top of a tower. It reaches the ground in 4 seconds. What is the height of the tower? (Assume g = 10 m/s²)
- A. 40 m
- B. 80 m
- C. 160 m
- D. 20 m
Answer: B. 80 m — Using s = ut + (1/2)gt², with u=0, g=10 m/s², t=4 s, we get s = 0*(4) + (1/2)*10*(4)² = 5 * 16 = 80 m.
MCQ 5. A particle starts from rest and accelerates such that its acceleration 'a' at time 't' is given by a = 3t + 2 m/s². What is its velocity after 2 seconds?
- A. 10 m/s
- B. 8 m/s
- C. 12 m/s
- D. 6 m/s
Answer: A. 10 m/s — Velocity is the integral of acceleration with respect to time. v = ∫ (3t + 2) dt = (3t²/2) + 2t + C. Since it starts from rest (v=0 at t=0), C=0. At t=2s, v = (3*2²/2) + 2*2 = 6 + 4 = 10 m/s.
MCQ 6. The slope of a position-time graph gives:
- A. Displacement
- B. Velocity
- C. Acceleration
- D. Speed
Answer: B. Velocity — The slope of a position-time (x-t) graph represents the rate of change of position, which is defined as velocity.
MCQ 7. A car travels the first half of a distance with a speed of 40 km/h and the second half with a speed of 60 km/h. What is the average speed of the car?
- A. 50 km/h
- B. 48 km/h
- C. 45 km/h
- D. 52 km/h
Answer: B. 48 km/h — For equal distances, the average speed formula is 2v₁v₂ / (v₁ + v₂). Here, v₁ = 40 km/h and v₂ = 60 km/h. So, average speed = (2 * 40 * 60) / (40 + 60) = 4800 / 100 = 48 km/h.
MCQ 8. A body is thrown vertically upwards with an initial velocity of 20 m/s. At what height will its velocity be half of its initial velocity? (Take g = 10 m/s²)
- A. 10 m
- B. 15 m
- C. 20 m
- D. 5 m
Answer: B. 15 m — Initial velocity u = 20 m/s. Final velocity v = u/2 = 10 m/s. Using v² = u² + 2as, with a = -g = -10 m/s². (10)² = (20)² + 2(-10)s. 100 = 400 - 20s, which gives 20s = 300, so s = 15 m.
MCQ 9. When an object is in uniform motion along a straight line, its velocity-time graph is a:
- A. Straight line inclined to the time axis
- B. Straight line parallel to the time axis
- C. Parabola
- D. Circle
Answer: B. Straight line parallel to the time axis — Uniform motion implies constant velocity and zero acceleration. Therefore, the velocity-time graph will be a horizontal straight line parallel to the time axis.
MCQ 10. A train starts from rest with a uniform acceleration of 0.5 m/s². What is its velocity after 100 m?
- A. 5 m/s
- B. 10 m/s
- C. 15 m/s
- D. 20 m/s
Answer: B. 10 m/s — Using v² = u² + 2as, with u=0, a=0.5 m/s², s=100 m. v² = 0² + 2*(0.5)*(100) = 100. So, v = √100 = 10 m/s.
MCQ 11. Two cars A and B are moving in the same direction with velocities 30 km/h and 45 km/h respectively. If car A is 10 km ahead of car B, how long will it take for car B to catch up with car A?
- A. 1 hour
- B. 2/3 hour
- C. 3/2 hour
- D. 1/2 hour
Answer: B. 2/3 hour — The relative velocity of car B with respect to car A is V_BA = V_B - V_A = 45 - 30 = 15 km/h. Car B needs to cover a relative distance of 10 km. Time = Distance / Relative Velocity = 10 km / 15 km/h = 2/3 hour.
MCQ 12. The area under a velocity-time graph represents:
- A. Speed
- B. Displacement
- C. Acceleration
- D. Time
Answer: B. Displacement — The area under the velocity-time (v-t) graph between two time instants gives the displacement of the object during that interval.
MCQ 13. An object travels 10 m in 4 seconds and then another 10 m in 2 seconds. What is the average speed of the object?
- A. 3.33 m/s
- B. 2.5 m/s
- C. 5 m/s
- D. 20 m/s
Answer: A. 3.33 m/s — Total distance = 10 m + 10 m = 20 m. Total time = 4 s + 2 s = 6 s. Average speed = Total distance / Total time = 20 m / 6 s = 10/3 m/s ≈ 3.33 m/s.
MCQ 14. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further will it penetrate before coming to rest, assuming constant deceleration?
- A. 1 cm
- B. 2 cm
- C. 3 cm
- D. 0.5 cm
Answer: A. 1 cm — Let initial velocity be u. For the first 3 cm: (u/2)² = u² + 2a(3) => u²/4 = u² + 6a => -3u²/4 = 6a => a = -u²/8. For further penetration (s'): 0² = (u/2)² + 2as' => 0 = u²/4 + 2(-u²/8)s' => 0 = u²/4 - (u²/4)s' => s' = 1 cm.
MCQ 15. Displacement is a vector quantity because it has:
- A. Magnitude only
- B. Direction only
- C. Both magnitude and direction
- D. No magnitude and no direction
Answer: C. Both magnitude and direction — A vector quantity is defined as having both magnitude and direction. Displacement represents the shortest distance from the initial to the final position, including the direction.
MCQ 16. A train is moving at a speed of 72 km/h. Brakes are applied and it undergoes uniform retardation of 2 m/s². How far will the train travel before it stops?
- A. 100 m
- B. 200 m
- C. 40 m
- D. 80 m
Answer: A. 100 m — Convert initial velocity to m/s: u = 72 km/h = 72 * (5/18) m/s = 20 m/s. Final velocity v = 0. Acceleration a = -2 m/s². Using v² = u² + 2as, we have 0² = (20)² + 2(-2)s. This gives 0 = 400 - 4s, so 4s = 400, and s = 100 m.
MCQ 17. A ball is dropped from a height h. It travels 9h/25 in the last second. What is the height h? (Take g = 10 m/s²)
- A. 125 m
- B. 100 m
- C. 50 m
- D. 200 m
Answer: A. 125 m — Let T be the total time of fall. h = (1/2)gT². The distance in the Tth second is s_T = (g/2)(2T-1). Given s_T = 9h/25. Substituting and solving for T: (g/2)(2T-1) = (9/25)(1/2)gT², which simplifies to 9T² - 50T + 25 = 0. The valid solution for T > 1s is T = 5s. Then h = (1/2)*10*(5)² = 125 m.
MCQ 18. The acceleration of a particle is constant. Its velocity-time graph will be a:
- A. Straight line with positive slope
- B. Straight line with negative slope
- C. Straight line parallel to the time axis
- D. Straight line with non-zero slope
Answer: D. Straight line with non-zero slope — For constant acceleration, the velocity changes uniformly with time, so the velocity-time graph is always a straight line. If the acceleration is non-zero, this line will have a non-zero slope (either positive or negative). If the acceleration is zero, the line would be parallel to the time axis. Option D best covers the general case of non-zero constant acceleration.
MCQ 19. A train starts from a station with uniform acceleration 'a₁', then moves with constant velocity for some time, and finally decelerates uniformly at 'a₂' to come to rest at the next station. If the total distance covered is 's' and the total time taken is 't', find the average velocity.
- A. s/t
- B. (a₁+a₂)/2 * t
- C. s * (a₁+a₂)/(2t)
- D. (a₁a₂ / (a₁+a₂)) * t
Answer: A. s/t — Average velocity is defined as total displacement divided by total time. Given that the total distance covered is 's' and total time taken is 't', the average velocity is simply s/t, assuming displacement equals distance in this straight-line motion.
MCQ 20. Which of the following is the correct definition of instantaneous velocity?
- A. The change in position over a specific time interval.
- B. The total distance covered divided by total time taken.
- C. The velocity of an object at a particular instant of time.
- D. The rate of change of speed.
Answer: C. The velocity of an object at a particular instant of time. — Instantaneous velocity refers to the velocity of an object at a specific moment or instant in time, defined as the limit of average velocity as the time interval approaches zero.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.