Class 11 : Physics (In English) – Chapter 2: Motion in a Straight Line
EXPLANATION & SUMMARY
🔶 1. Introduction
This chapter deals with motion in one dimension, i.e., along a straight line. It’s part of kinematics, the branch of mechanics that describes motion without regard to its causes.
🔶 2. Rest and Motion
A body is at rest if it does not change its position with time relative to a reference point.
A body is in motion if it changes its position with time.
Rest and motion are relative concepts depending on the observer’s frame of reference.
🔶 3. Position, Path Length, and Displacement
Position: Location of a particle at a given time, measured from a reference point (usually origin).
Path length: The actual length of the path traveled (scalar quantity).
Displacement: Change in position in a straight line with direction (vector quantity).
Example: For a round trip, displacement = 0 but path length > 0.
🔶 4. Speed and Velocity
Speed = Distance / Time (scalar)
Velocity = Displacement / Time (vector)
Types:
Average speed = Total distance / Total time
Average velocity = Total displacement / Total time
Instantaneous velocity = Limit of average velocity as time interval → 0
That is: v = dx/dt
🔶 5. Acceleration
Acceleration is the rate of change of velocity with time:
a = dv/dt
Positive acceleration: speed increases
Negative acceleration (retardation): speed decreases
Zero acceleration: uniform velocity
🔶 6. Uniform and Non-uniform Motion
Uniform motion: Constant velocity, zero acceleration
Uniformly accelerated motion: Acceleration is constant
Non-uniform motion: Variable velocity and/or variable acceleration
🔶 7. Equations of Motion for Uniform Acceleration
Let:
u = initial velocity
v = final velocity
a = acceleration
s = displacement
t = time
Then:
v = u + at
s = ut + (1/2)at²
v² = u² + 2as
These are valid only when a is constant.
🔶 8. Graphical Representation
(a) Position-Time Graph
Slope = velocity
Straight line = constant velocity
Curved line = acceleration
(b) Velocity-Time Graph
Slope = acceleration
Area under curve = displacement
(c) Acceleration-Time Graph
Area under curve = change in velocity
🔶 9. Free Fall
A body falling freely under gravity (no air resistance)
Acceleration = g = 9.8 m/s² downward
Use the same kinematic equations with a = ±g
Upward motion: a = -g
Downward motion: a = +g
🔶 10. Relative Velocity
If two objects A and B are moving with velocities vA and vB:
Velocity of A with respect to B:
vAB = vA – vB
Velocity of B with respect to A:
vBA = vB – vA
🔶 11. Motion Under Gravity (Vertical Motion)
Upward motion (retardation):
v = u – gt
s = ut – (1/2)gt²
v² = u² – 2gs
Downward motion (acceleration):
v = u + gt
s = ut + (1/2)gt²
v² = u² + 2gs
At highest point: v = 0
🔶 12. Sign Conventions
Motion in chosen direction → positive (+)
Motion in opposite direction → negative (−)
Use consistent coordinate axes in all calculations
🔶 13. Calculus in Motion
v = dx/dt → Velocity is derivative of position
a = dv/dt → Acceleration is derivative of velocity
x = ∫v dt → Position is integral of velocity
v = ∫a dt → Velocity is integral of acceleration
🔶 14. Sample Numerical Application
Example: A particle starts with u = 5 m/s and accelerates uniformly at a = 2 m/s² for t = 3 s.
Find:
Final velocity: v = u + at = 5 + 2×3 = 11 m/s
Displacement: s = ut + (1/2)at² = 5×3 + 0.5×2×9 = 15 + 9 = 24 m
📘 PART B: CRISP SUMMARY (~300 WORDS)
🔷 Summary – Motion in a Straight Line
Motion along a straight line is called rectilinear motion.
It is described using displacement, velocity, and acceleration.
Displacement is the vector difference in position.
Path length is the total distance traveled, always positive.
Velocity is the rate of change of displacement.
Acceleration is the rate of change of velocity.
For uniform acceleration, use these three equations:
v = u + at
s = ut + (1/2)at²
v² = u² + 2as
Free fall is motion under gravity with a = g = 9.8 m/s².
Use positive and negative signs to indicate direction.
Graphs are helpful tools to analyze motion:
Slope of position-time graph = velocity
Slope of velocity-time graph = acceleration
Area under v-t graph = displacement
Relative velocity gives one object’s velocity as seen from another.
Using derivatives and integrals, we can find:
v = dx/dt
a = dv/dt
x = ∫v dt
v = ∫a dt
This chapter builds a strong foundation for understanding motion before progressing to two or three dimensions.
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QUESTIONS FROM TEXTBOOK
🔷 Question 2.1
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
✅ Answer:
(a) ✅ Yes, because the size of the railway carriage is negligible compared to the distance between stations.
(b) ✅ Yes, if the size of the monkey is negligible compared to the circular track.
(c) ❌ No, rotation of the ball is significant and cannot be neglected.
(d) ❌ No, the beaker undergoes rotational motion and deformation during the fall.
🔷 Question 2.2
The position-time (x–t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) (A and B / B) reach home at the same/different time
(e) (A/B) overtakes (B/A) on the road (once/twice).
✅ Answer (based on Fig. 2.9):
(a) A lives closer to school than B
(b) B starts from school earlier than A
(c) A walks faster than B (steeper slope)
(d) A and B reach home at the same time
(e) A overtakes B once
🔷 Question 2.3
A woman starts from her home at 9:00 am, walks with a speed of 5 km h⁻¹ on a straight road up to her office 2.5 km away, stays at the office up to 5:00 pm, and returns home by an auto with a speed of 25 km h⁻¹. Choose suitable scales and plot the x–t graph of her motion.
✅ Answer:
📌 From 9:00 am to 9:30 am → walking to office
📌 From 9:30 am to 5:00 pm → stationary at office
📌 From 5:00 pm to 5:06 pm → returning by auto
Graph Features:
Straight inclined line (walk)
Horizontal line (stay at office)
Steep declining line (return by auto)
[Graph is not to be drawn here as per plain text instruction.]
🔷 Question 2.4
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x–t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
✅ Answer:
Each cycle of motion (5 steps forward, 3 steps back):
📌 Net displacement = 2 m
📌 Time per cycle = 8 s
To cover first 10 m (from cycles):
📌 2 m per 8 s ⇒ 5 cycles = 10 m in 40 s
Remaining:
📌 Now, move forward 3 steps = 3 m in 3 s ⇒ reach 13 m in 43 s
✅ Final Answer: Drunkard takes 43 seconds to fall into the pit.
🔷 Question 2.5
A car moving along a straight highway with speed of 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation (assumed uniform), and how long does the car take to stop?
✅ Answer:
📌 u = 126 km/h = 35 m/s
📌 v = 0, s = 200 m
📘 Use v² = u² + 2as
0 = (35)² + 2a(200)
a = –(1225 / 400) = –3.06 m/s²
📘 Time, t = (v – u)/a = (0 – 35)/–3.06 ≈ 11.44 s
✅ Final Answer:
✔️ Retardation = 3.06 m/s²
✔️ Time to stop = 11.44 s
🔷 Question 2.6
A player throws a ball upwards with an initial speed of 29.4 m s⁻¹.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point?
(c) Choose the x = 0 and t = 0 to be the location and time of the ball at its highest point, vertically downward direction to be positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands?
✅ Answer:
(a) Acceleration is downward in all cases = –9.8 m/s²
(b)
📌 At highest point:
Velocity = 0, Acceleration = –9.8 m/s²
(c)
📌 Upward motion (before reaching x = 0):
Position: negative, Velocity: negative, Acceleration: positive
📌 Downward motion (after crossing x = 0):
Position: positive, Velocity: positive, Acceleration: positive
(d)
📘 u = 29.4 m/s, a = –9.8 m/s²
Max height (H) = u²/2g = (29.4)² / (2 × 9.8) = 882.36 / 19.6 = 45 m
Time to rise = u/g = 29.4 / 9.8 = 3 s
Total time = 2 × 3 = 6 s
✅ Final Answers:
✔️ Height = 45 m
✔️ Time to return = 6 s
🔷 Question 2.7
Read each statement below carefully and state with reasons and examples, if it is true or false:
(a) A particle in one-dimensional motion with zero speed at an instant may have non-zero acceleration at that instant.
(b) A particle in one-dimensional motion with zero speed may have non-zero velocity.
(c) A particle in one-dimensional motion with constant speed must have zero acceleration.
(d) A particle in one-dimensional motion with positive value of acceleration must be speeding up.
✅ Answer:
(a) ✅ True – At turning point in projectile motion, speed is 0 but acceleration ≠ 0.
(b) ❌ False – Zero speed means zero magnitude of velocity.
(c) ✅ True – Constant speed ⇒ no change in velocity ⇒ zero acceleration.
(d) ❌ False – Positive acceleration may be opposite to velocity → object may slow down.
🔷 Question 2.8
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.
✅ Answer:
📌 Initial velocity u = 0, height = 90 m
Use: h = (1/2)gt² ⇒ t = √(2h/g) = √(2 × 90 / 9.8) ≈ 4.29 s to fall down
📘 On hitting floor:
Speed just before impact = √(2gh) = √(2 × 9.8 × 90) ≈ 42 m/s
After first rebound = 0.9 × 42 = 37.8 m/s
Next rebound = 0.9 × 37.8 = 34.02 m/s
… and so on
⛔ Graph involves changes in slope and values over time, showing decreasing peaks.
Since diagrams are not allowed here, just note:
✅ Speed drops 10% after each bounce
✅ Graph is zigzag with decreasing peaks
🔷 Question 2.9
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval.
✅ Answer:
(a)
📘 Magnitude of displacement is the shortest distance between initial and final positions.
📘 Total path length is the actual distance travelled.
🟢 Example:
If a person walks 3 m east, then 4 m west:
Displacement = 1 m
Path length = 7 m
(b)
📘 Average velocity = displacement / time
📘 Average speed = total path / time
🟢 Example: Using same case above, time = 7 s
Avg velocity = 1/7 m/s
Avg speed = 7/7 = 1 m/s
🔷 Question 2.10
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h⁻¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h⁻¹. What is
(a) the magnitude of average velocity, and
(b) the average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
✅ Answer:
📘 (i) Time to go = 2.5 / 5 = 0.5 h = 30 min
📘 (ii) Time to return = 2.5 / 7.5 = 1/3 h ≈ 20 min
Total round trip = 50 min = 5/6 h
(a)
Only forward trip (30 min):
Displacement = 2.5 km
Time = 0.5 h
⇒ Avg velocity = 2.5 / 0.5 = 5 km/h
(b)
(i) 0–30 min:
Avg speed = 2.5 / 0.5 = 5 km/h
(ii) 0–50 min:
Total path = 5 km
Time = 5/6 h
⇒ Avg speed = 5 / (5/6) = 6 km/h
(iii) 0–40 min:
📘 Forward = 2.5 km in 30 min
📘 Return = 10 min at 7.5 km/h = 1.25 km
Displacement = 2.5 − 1.25 = 1.25 km
Total path = 2.5 + 1.25 = 3.75 km
Time = 2/3 h
Avg velocity = 1.25 / (2/3) = 1.875 km/h
Avg speed = 3.75 / (2/3) = 5.625 km/h
🔷 Question 2.11
In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
✅ Answer:
📘 Instantaneous speed = |Instantaneous velocity|
This is because at a particular instant, displacement and distance become the same (no path bending at zero time interval).
Therefore, direction doesn’t matter and both values are identical.
🔷 Question 2.12
Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
✅ Answer:
(a) ✅ Possible (straight line, uniform velocity)
(b) ✅ Possible (circular path projection on x)
(c) ❌ Not possible – particle can’t have more than one x for the same time
(d) ❌ Not possible – speed can’t be negative
✅ Graphs (c) and (d) are not valid representations of one-dimensional motion.
🔷 Question 2.13
Figure 2.11 shows the x–t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
✅ Answer:
No, this interpretation is incorrect because the x–t graph only shows position vs. time.
📌 It does not represent the path or trajectory (e.g., straight or curved).
📌 The graph indicates non-uniform motion with increasing velocity for t > 0.
Suitable context:
An object accelerating uniformly from rest, like a ball rolling down an inclined plane.
🔷 Question 2.14
A police van moving on a highway with a speed of 30 km h⁻¹ fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h⁻¹. If the muzzle speed of the bullet is 150 m s⁻¹, with what speed does the bullet hit the thief’s car?
(Note: Obtain that speed which is relevant for damaging the thief’s car).
✅ Answer:
📌 Convert all to m/s:
Police van = 30 km/h = 8.33 m/s
Thief’s car = 192 km/h = 53.33 m/s
Bullet muzzle speed (relative to van) = 150 m/s
📘 Bullet speed w.r.t ground = 150 + 8.33 = 158.33 m/s
📘 Relative speed of bullet w.r.t thief’s car =
158.33 − 53.33 = 105 m/s
✅ Answer: The bullet hits the thief’s car with speed = 105 m/s
🔷 Question 2.15
Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
✅ Answer:
(a) Displacement-time graph with a break at point B:
🟢 Suggests the object moves with constant velocity (uniform motion) from O to A, then stops for a while (horizontal line AB), and then resumes uniform motion (positive slope again).
✅ Physical Situation: A person walks at constant speed, pauses to talk on the phone (at point B), then resumes walking.
(b) Velocity-time graph with repeated sharp reversals in direction:
🟢 Velocity alternates between positive and negative with same magnitude.
✅ Physical Situation: A ball bouncing vertically between floor and ceiling with constant speed but instant reversal at each collision.
(c) Acceleration-time graph with sharp peak:
🟢 High acceleration for a very short time followed by zero acceleration.
✅ Physical Situation: A bullet fired from a gun – large force (acceleration) for a brief time, then uniform motion.
🔷 Question 2.16
Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13.) Give the signs of position, velocity and acceleration variables of the particle at
t = 0.3 s, 1.2 s, 2 s, –1.2 s.
✅ Answer:
🟢 Position → based on location above/below x-axis
🟢 Velocity → slope of tangent to x–t curve
🟢 Acceleration → opposite to displacement (restoring force)
Time (s) Position Velocity Acceleration
0.3 + + –
1.2 0 – 0
2.0 – – +
–1.2 0 + 0
🔷 Question 2.17
Figure 2.14 gives the x–t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
✅ Answer:
📘 Average speed = total path / time
📘 Average velocity = displacement / time
🟢 From graph:
Interval 1: Moderate slope
Interval 2: Steeper slope → highest average speed
Interval 3: Flat slope → lowest average speed
✅ Greatest average speed: Interval 2
✅ Least average speed: Interval 3
✅ Average velocity:
Interval 1: Positive
Interval 2: Positive
Interval 3: Zero (no displacement)
🔷 Question 2.18
Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choose the positive direction as the constant direction of motion. Give the signs of a in the three intervals. What are the accelerations at points A, B, C and D?
✅ Answer:
📘 Acceleration = (change in speed) / (time)
🟢 From graph:
Interval 1 (rising): Speed increases → positive acceleration
Interval 2 (falling): Speed decreases rapidly → negative acceleration
Interval 3 (small bump): Smaller increase → lower acceleration
✅ Greatest magnitude of average acceleration: Interval 2 (falling steeply)
✅ Greatest average speed: Interval 1 (highest area under curve)
✅ Signs of acceleration:
Interval 1: Positive
Interval 2: Negative
Interval 3: Positive
✅ Acceleration at points:
A: Positive (slope up)
B: Zero (top point)
C: Negative (slope down)
D: Zero (valley point)
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OTHER IMPORTANT QUESTIONS FOR EXAMS
(CBSE MODEL QUESTIONS PAPER)
ESPECIALLY MADE FROM THIS LESSON ONLY
Q1. Which of the following is a scalar quantity?
(A) Displacement
(B) Velocity
(C) Speed
(D) Acceleration
Answer: (C) Speed
Q2. The slope of a position-time graph gives:
(A) Acceleration
(B) Speed
(C) Displacement
(D) Velocity
Answer: (D) Velocity
Q3. A body is said to be in uniform motion if:
(A) Its velocity changes at constant rate
(B) It covers equal distances in unequal intervals
(C) It covers unequal distances in equal intervals
(D) It covers equal distances in equal intervals
Answer: (D) It covers equal distances in equal intervals
Q4. A car moves with a uniform acceleration of 2 m/s². Its speed after 5 seconds is:
(A) 10 m/s
(B) 5 m/s
(C) 20 m/s
(D) 2 m/s
Answer: (A) 10 m/s
(Since v = u + at, with u = 0, v = 2 × 5 = 10 m/s)
Q5. Which of the following represents non-uniform acceleration?
(A) A ball falling under gravity
(B) A vehicle moving with constant speed
(C) A rocket launching upward
(D) A stone thrown vertically upward in vacuum
Answer: (C) A rocket launching upward
Q6. SI unit of displacement is:
(A) cm
(B) m
(C) km
(D) mm
Answer: (B) m
Q7. Assertion (A): A body having zero velocity can still have acceleration.
Reason (R): Acceleration is the rate of change of velocity, not velocity itself.
(A) Both A and R are true, and R is the correct explanation of A
(B) Both A and R are true, but R is not the correct explanation of A
(C) A is true, R is false
(D) A is false, R is true
Answer: (A) Both A and R are true, and R is the correct explanation of A
Q8. Displacement of a body is zero. Which of the following is always true?
(A) Distance travelled is zero
(B) Final position is equal to initial position
(C) The body is at rest
(D) Velocity is constant
Answer: (B) Final position is equal to initial position
Q9. A particle travels 20 m north, then 15 m south. What is its net displacement?
(A) 35 m
(B) 5 m
(C) 25 m
(D) 0 m
Answer: (B) 5 m
Q10. Write the relation between instantaneous speed and magnitude of velocity.
Answer: Instantaneous speed is equal to the magnitude of instantaneous velocity.
Q11. What does the area under a velocity-time graph represent?
Answer: Displacement of the particle.
Q12. Define average speed.
Answer: Average speed = Total distance travelled / Total time taken
Q13. A train starts from rest. What is its initial velocity?
Answer: Zero (0 m/s)
Q14. What is the value of displacement when a particle returns to its starting point?
Answer: Zero
Q15. Write the first equation of motion and define each term.
Answer: v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Q16. What is the shape of the velocity-time graph for uniformly accelerated motion?
Answer: A straight line with constant slope
Q17. Case-based MCQ:
A car accelerates from rest with a constant acceleration and reaches a velocity of 20 m/s in 10 seconds.
What is the displacement of the car in these 10 seconds?
(A) 100 m
(B) 200 m
(C) 150 m
(D) 50 m
Answer: (B) 200 m
(Use: s = ut + ½at², u = 0, a = 2 m/s² ⇒ s = ½ × 2 × 10² = 100 m)
Correction: First, find acceleration: a = (v – u)/t = (20 – 0)/10 = 2 m/s²
Then s = ut + ½at² = 0 + ½ × 2 × 100 = 100 m
Correct Answer: (A) 100 m
Q18. Which of the following statements is true?
(A) Displacement is always positive
(B) Displacement can be positive, negative, or zero
(C) Displacement is always more than distance
(D) Displacement and distance are always equal
Answer: (B) Displacement can be positive, negative, or zero
Section B: Q19–Q23 (2 Marks Each)
Q19. A particle moves in a straight line with a velocity of 5 m/s for 4 s and then with 3 m/s for 2 s in the same direction. Calculate the average velocity.
Answer:
Total displacement = (5 × 4) + (3 × 2) = 20 + 6 = 26 m
Total time = 4 + 2 = 6 s
Average velocity = Total displacement / Total time = 26 / 6 = 4.33 m/s
Q20. Write two differences between distance and displacement.
Answer:
Distance is a scalar quantity, while displacement is a vector quantity.
Distance is always positive; displacement can be positive, negative, or zero.
Q21. A car accelerates from rest at 2 m/s². Find its speed after 8 seconds and distance covered in that time.
Answer:
Given: u = 0, a = 2 m/s², t = 8 s
v = u + at = 0 + 2 × 8 = 16 m/s
s = ut + ½at² = 0 + ½ × 2 × 64 = 64 m
Q22. A train is moving with a velocity of 36 km/h. How much distance will it cover in 25 minutes?
Answer:
36 km/h = 10 m/s
Time = 25 min = 1500 s
Distance = speed × time = 10 × 1500 = 15000 m = 15 km
Q23. Derive the second equation of motion: s = ut + ½at² using calculus method.
Answer:
We know: a = dv/dt
Integrate: v = u + at
Now, dx/dt = v = u + at
Integrate again:
∫dx = ∫(u + at)dt
x = ut + ½at²
∴ Displacement s = x = ut + ½at²
Section C: Q24–Q28 (3 Marks Each)
Q24. A car starts from rest, accelerates uniformly at 3 m/s² for 5 seconds, then moves with constant speed for 10 seconds and finally retards at 2 m/s² till it stops. Find the total distance covered.
Answer:
Stage 1 (acceleration):
u = 0, a = 3, t = 5
v = u + at = 0 + 3×5 = 15 m/s
s₁ = ut + ½at² = 0 + ½×3×25 = 37.5 m
Stage 2 (uniform):
v = 15 m/s, t = 10 s
s₂ = vt = 15 × 10 = 150 m
Stage 3 (retardation):
v = 15 m/s, a = –2, v_f = 0
v² = u² + 2as
0 = 225 – 4s ⇒ s₃ = 56.25 m
Total distance = s₁ + s₂ + s₃ = 37.5 + 150 + 56.25 = 243.75 m
Q25. A particle starts from origin and moves along x-axis such that its velocity varies with time as v(t) = 3t². Find displacement in 2 seconds.
Answer:
v = dx/dt = 3t²
Integrate: dx = 3t² dt
x = ∫₀² 3t² dt = 3∫₀² t² dt = 3 × [t³/3]₀² = [t³]₀² = 8 – 0 = 8 m
Q26. Two cars A and B start from the same point. Car A moves with constant speed of 10 m/s, while car B starts from rest and accelerates at 2 m/s². When and where will they meet?
Answer:
Let time = t
Distance by A = 10t
Distance by B = ½ × 2 × t² = t²
Equating: 10t = t² ⇒ t² – 10t = 0 ⇒ t(t – 10) = 0 ⇒ t = 10 s
Distance = 10 × 10 = 100 m
Q27. A particle moves along a straight line. Its displacement x at time t is given by x = 2t + 3t². Find velocity and acceleration at t = 2 s.
Answer:
x = 2t + 3t²
v = dx/dt = 2 + 6t ⇒ v(2) = 2 + 6×2 = 14 m/s
a = dv/dt = 6 ⇒ 6 m/s²
Q28. A stone is dropped from the top of a tower 80 m high. Find the time taken to reach the ground and final velocity.
Answer:
Given: u = 0, h = 80 m, g = 9.8 m/s²
Using s = ut + ½gt²
80 = 0 + ½ × 9.8 × t² ⇒ t² = 160 / 9.8 ≈ 16.33 ⇒ t ≈ 4.04 s
v = u + gt = 0 + 9.8 × 4.04 ≈ 39.6 m/s
Section D: Q29–Q31 (4 Marks Each)
Case-Based Questions with Internal Parts
Q29.
Read the following case and answer the questions:
A bus starts from rest and moves with uniform acceleration. A student standing nearby records its velocity at various times and plots a velocity-time graph.
(a) What physical quantity is represented by the slope of the velocity-time graph?
(b) What does the area under this graph represent?
(c) If the acceleration is 2 m/s², find velocity after 6 seconds.
(d) Find the displacement in this time.
Answer:
(a) Slope of velocity-time graph = Acceleration
(b) Area under the graph = Displacement
(c) v = u + at = 0 + 2 × 6 = 12 m/s
(d) s = ut + ½at² = 0 + ½ × 2 × 36 = 36 m
Q30.
Read the following case and answer:
A stone is projected vertically upward with a speed of 20 m/s from the top of a building 45 m high.
(a) How much time will it take to reach the highest point?
(b) What is the maximum height from the ground?
(c) What will be the velocity just before it strikes the ground?
(d) What is the total time of flight?
Answer:
Given: u = 20 m/s, h = 45 m, g = 9.8 m/s²
(a) Time to reach max height: t = u/g = 20 / 9.8 ≈ 2.04 s
(b) Height above point of projection = u²/2g = 400 / 19.6 ≈ 20.41 m
∴ Maximum height from ground = 45 + 20.41 = 65.41 m
(c) Use: v² = u² + 2gh = 0 + 2×9.8×65.41 ≈ v² = 1280 ⇒ v ≈ 35.78 m/s
(d) Total time: First up = 2.04 s
Time down: h = 65.41, use h = ½gt² ⇒ t² = 2×65.41 / 9.8 ≈ 13.34 ⇒ t = 3.65 s
Total time = 2.04 + 3.65 = 5.69 s
Q31.
Read the case and answer:
Two friends A and B start from the same point. A walks at 5 m/s, and B runs at 10 m/s but starts 10 seconds after A.
(a) How far would A have gone before B starts?
(b) After how much time will B catch up with A?
(c) How far will they be from the starting point when they meet?
(d) Who has travelled more distance by that time?
Answer:
(a) Distance by A in 10 s = 5 × 10 = 50 m
(b) Let t = time after B starts
In t seconds,
Distance by A = 5(t + 10)
Distance by B = 10t
Equating: 5t + 50 = 10t ⇒ 50 = 5t ⇒ t = 10 s
(c) Distance = 10 × 10 = 100 m
(d) A travels = 5 × 20 = 100 m, B travels = 10 × 10 = 100 m
Both travel same distance
Section E: Q32–Q35 (5 Marks Each)
Long Answer Theory or Numerical Questions (with Full Steps)
Q32.
Derive all three equations of motion using graphical method.
Answer:
Graph: Velocity-time graph of uniformly accelerated motion is a straight line.
Let initial velocity = u, final velocity = v, time = t, acceleration = a
First equation:
Slope of line = a = (v – u) / t ⇒ v = u + at
Second equation:
Displacement = Area under v–t graph = Area of trapezium
s = ½ × (u + v) × t
Using v = u + at ⇒ s = ut + ½at²
Third equation:
From s = ½(u + v)t
Eliminate t from v = u + at ⇒ t = (v – u)/a
s = ½(u + v) × (v – u)/a
s = (v² – u²)/(2a) ⇒ v² = u² + 2as
Q33.
A car moves along a straight road for 5 s with constant acceleration, then 10 s at constant speed and finally 5 s with retardation. Draw v–t and x–t graphs. Also calculate total displacement if maximum velocity was 20 m/s.
Answer:
Let:
Stage 1: a = ?, t₁ = 5 s, u = 0 ⇒ v = u + at = 20 ⇒ a = 4 m/s²
Displacement s₁ = ut + ½at² = ½ × 4 × 25 = 50 m
Stage 2: v = 20 m/s, t₂ = 10 s ⇒ s₂ = vt = 20 × 10 = 200 m
Stage 3: Retardation, t₃ = 5 s, final v = 0
Use v = u – at ⇒ 0 = 20 – a × 5 ⇒ a = 4 m/s²
s₃ = ut – ½at² = 20 × 5 – ½ × 4 × 25 = 100 – 50 = 50 m
Total displacement = 50 + 200 + 50 = 300 m
(Students are expected to draw trapezium-shaped velocity-time graph and a curve-linear x–t graph showing increase, constant slope, and decrease)
Q34.
A car starts from rest and accelerates at 2 m/s² for 10 seconds, then continues at constant speed for 5 s, and finally decelerates to rest in 5 s.
Find:
(a) Total distance travelled
(b) Time at which maximum speed occurs
(c) Plot rough v–t graph
(d) Calculate average speed
Answer:
Stage 1:
u = 0, a = 2 m/s², t = 10 s
v = u + at = 20 m/s
s₁ = ut + ½at² = 100 m
Stage 2:
Constant speed = 20 m/s, t = 5 s
s₂ = 20 × 5 = 100 m
Stage 3:
v = 20 m/s, final v = 0, t = 5 s
Retardation a = 20/5 = 4 m/s²
s₃ = ut – ½at² = 20×5 – ½×4×25 = 100 – 50 = 50 m
Total distance = 100 + 100 + 50 = 250 m
Total time = 20 s ⇒ Average speed = 250 / 20 = 12.5 m/s
Q35.
A body travels 15 m in first 3 s and 25 m in next 3 s along a straight line with uniform acceleration. Find:
(a) Initial velocity
(b) Acceleration
(c) Velocity at end of 6 s
Answer:
Let u = initial velocity, a = acceleration
First 3 s: s₁ = 15 = ut + ½at² = 3u + 4.5a — (1)
Next 3 s: s₂ = 25 = u’×3 + ½a×9
Where u’ = u + 3a ⇒
s₂ = (u + 3a)×3 + 4.5a = 3u + 9a + 4.5a = 3u + 13.5a — (2)
Subtract (1):
(2) – (1): 25 – 15 = (3u + 13.5a) – (3u + 4.5a) = 9a
⇒ 10 = 9a ⇒ a = 10/9 m/s²
From (1): 15 = 3u + 4.5 × (10/9) ⇒ 3u = 15 – 5 = 10 ⇒ u = 10/3 m/s
v₆ = u + at = 10/3 + (10/9)×6 = 10/3 + 20/3 = 30/3 = 10 m/s
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