This MCQ module is based on: Position Velocity
TOPIC 5 OF 33
Position Velocity
🎓 Class 11
Physics
CBSE
Theory
Ch 2 – Motion in a Straight Line
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Position Velocity
Upload images, PDFs, or Word documents to include their content in assessment generation.
Position Velocity
2.1 Introduction
Motion is one of the most common phenomena around us. People walking, vehicles on roads, the flow of blood through arteries, the orbit of the Earth around the Sun — all are examples of bodies in motion. The branch of physics that deals with describing motion without investigating its causes is called kinematics.
In this chapter, we focus on motion along a straight line — the simplest type of motion. An object is treated as a point object when its size is negligibly small compared to the distance it moves. For instance, a car covering hundreds of kilometres can be treated as a point, while a train entering a 50-metre platform cannot (its length is comparable to the distance).
2.2 Position, Path Length and Displacement
To describe the motion of an object along a straight line, we first choose a convenient point as the origin (O) and define a positive direction. The position of the object is then its distance from the origin, with a sign indicating direction.
Path Length (Distance)
Path length is the total distance covered by the object along its actual route. It is a scalar (no direction) and is always non-negative.
Displacement
Displacement is the change in position of the object:
\[\Delta x = x_2 - x_1\]
Displacement is a vector — it can be positive, negative, or zero. A crucial relationship links displacement and path length:
Key Inequality: The magnitude of displacement is always less than or equal to the path length: \(|\Delta x| \le \text{path length}\). Equality holds only when the object moves in a single direction without reversing.
Worked Example 1: Distance vs Displacement
A jogger runs 400 m due east and then 300 m due west. Find (a) the path length, and (b) the displacement.
Solution:
Taking east as positive direction:
(a) Path length = total distance covered = 400 + 300 = \(\boxed{700 \text{ m}}\)
(b) Displacement = final position − initial position
Final position from origin = 400 − 300 = +100 m
\[\Delta x = +100 - 0 = \boxed{+100 \text{ m (east)}}\]
Notice: path length (700 m) > |displacement| (100 m), confirming \(|\Delta x| \le \text{path length}\).
Taking east as positive direction:
(a) Path length = total distance covered = 400 + 300 = \(\boxed{700 \text{ m}}\)
(b) Displacement = final position − initial position
Final position from origin = 400 − 300 = +100 m
\[\Delta x = +100 - 0 = \boxed{+100 \text{ m (east)}}\]
Notice: path length (700 m) > |displacement| (100 m), confirming \(|\Delta x| \le \text{path length}\).
2.3 Average Velocity and Average Speed
Average velocity is defined as the displacement per unit time:
\[\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}\]
Average speed is the total path length divided by total time:
\[\text{Average speed} = \frac{\text{Total path length}}{\Delta t}\]
Key Difference: Average velocity can be zero or negative; average speed is always positive. Also, average speed ≥ |average velocity|.
Worked Example 2 (NCERT Example 2.1): Car Journey
A car moves along a straight road. It covers 360 m from O to A in 18 minutes, then reverses and travels 120 m from A to B in 6 minutes. Find (a) average velocity, and (b) average speed for the entire trip.
Solution:
Given: O is the origin. Position of A = +360 m. The car reverses 120 m, so position of B = 360 − 120 = +240 m.
Total time = 18 + 6 = 24 min = 24 × 60 = 1440 s
(a) Average velocity: \[\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_B - x_O}{\Delta t} = \frac{240 - 0}{1440} = \frac{240}{1440}\] \[\boxed{\bar{v} = 0.167 \text{ m/s} \approx \frac{1}{6} \text{ m/s}}\]
(b) Average speed: \[\text{Average speed} = \frac{\text{Total path length}}{\Delta t} = \frac{360 + 120}{1440} = \frac{480}{1440}\] \[\boxed{\text{Average speed} = 0.333 \text{ m/s} = \frac{1}{3} \text{ m/s}}\]
As expected, average speed (0.333 m/s) > average velocity (0.167 m/s).
Given: O is the origin. Position of A = +360 m. The car reverses 120 m, so position of B = 360 − 120 = +240 m.
Total time = 18 + 6 = 24 min = 24 × 60 = 1440 s
(a) Average velocity: \[\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_B - x_O}{\Delta t} = \frac{240 - 0}{1440} = \frac{240}{1440}\] \[\boxed{\bar{v} = 0.167 \text{ m/s} \approx \frac{1}{6} \text{ m/s}}\]
(b) Average speed: \[\text{Average speed} = \frac{\text{Total path length}}{\Delta t} = \frac{360 + 120}{1440} = \frac{480}{1440}\] \[\boxed{\text{Average speed} = 0.333 \text{ m/s} = \frac{1}{3} \text{ m/s}}\]
As expected, average speed (0.333 m/s) > average velocity (0.167 m/s).
Worked Example 3: Two-Part Journey at Different Speeds
A cyclist covers the first half of a distance at 20 km/h and the second half at 30 km/h. Find the average speed for the whole journey.
Solution:
Let total distance = \(2d\). Each half = \(d\).
Time for first half: \(t_1 = \frac{d}{20}\)
Time for second half: \(t_2 = \frac{d}{30}\)
Total time: \(t = t_1 + t_2 = \frac{d}{20} + \frac{d}{30} = d\left(\frac{3+2}{60}\right) = \frac{5d}{60} = \frac{d}{12}\)
\[\text{Average speed} = \frac{2d}{d/12} = 2d \times \frac{12}{d} = \boxed{24 \text{ km/h}}\]
Note: The average speed (24 km/h) is NOT the arithmetic mean (25 km/h). For equal distances at different speeds, the average speed is the harmonic mean: \(\frac{2v_1 v_2}{v_1 + v_2} = \frac{2 \times 20 \times 30}{50} = 24\) km/h.
Let total distance = \(2d\). Each half = \(d\).
Time for first half: \(t_1 = \frac{d}{20}\)
Time for second half: \(t_2 = \frac{d}{30}\)
Total time: \(t = t_1 + t_2 = \frac{d}{20} + \frac{d}{30} = d\left(\frac{3+2}{60}\right) = \frac{5d}{60} = \frac{d}{12}\)
\[\text{Average speed} = \frac{2d}{d/12} = 2d \times \frac{12}{d} = \boxed{24 \text{ km/h}}\]
Note: The average speed (24 km/h) is NOT the arithmetic mean (25 km/h). For equal distances at different speeds, the average speed is the harmonic mean: \(\frac{2v_1 v_2}{v_1 + v_2} = \frac{2 \times 20 \times 30}{50} = 24\) km/h.
2.4 Instantaneous Velocity and Speed
The average velocity gives us a broad picture over a time interval. But what about the velocity at a specific instant? The instantaneous velocity is defined as:
\[v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}\]
Graphically, the instantaneous velocity at any point on an \(x\)-\(t\) graph equals the slope of the tangent to the curve at that point.
The instantaneous speed is simply the magnitude of instantaneous velocity: speed = \(|v|\).
Reading an x-t Graph
| x-t Graph Feature | Motion Interpretation |
|---|---|
| Straight line with positive slope | Uniform velocity in positive direction |
| Straight line with negative slope | Uniform velocity in negative direction |
| Horizontal line (zero slope) | Object at rest |
| Curve bending upward (increasing slope) | Positive acceleration (speeding up) |
| Curve bending downward (decreasing slope) | Negative acceleration (slowing down) |
Worked Example 4: Finding Velocity from Position Function
The position of a particle is given by \(x(t) = 3t^2 + 5t + 2\) (in metres, with \(t\) in seconds). Find (a) the velocity at \(t = 2\) s, and (b) the average velocity between \(t = 1\) s and \(t = 3\) s.
Solution:
(a) Instantaneous velocity = \(\frac{dx}{dt}\)
\[v = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5\] At \(t = 2\) s: \[v = 6(2) + 5 = 12 + 5 = \boxed{17 \text{ m/s}}\]
(b) Average velocity between \(t = 1\) s and \(t = 3\) s:
\[x(1) = 3(1)^2 + 5(1) + 2 = 10 \text{ m}\] \[x(3) = 3(9) + 5(3) + 2 = 27 + 15 + 2 = 44 \text{ m}\] \[\bar{v} = \frac{x(3) - x(1)}{3 - 1} = \frac{44 - 10}{2} = \boxed{17 \text{ m/s}}\]
Interesting: Here, the average velocity equals the instantaneous velocity at the midpoint (\(t = 2\)). This happens because acceleration is uniform (constant \(a = 6\) m/s²).
(a) Instantaneous velocity = \(\frac{dx}{dt}\)
\[v = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5\] At \(t = 2\) s: \[v = 6(2) + 5 = 12 + 5 = \boxed{17 \text{ m/s}}\]
(b) Average velocity between \(t = 1\) s and \(t = 3\) s:
\[x(1) = 3(1)^2 + 5(1) + 2 = 10 \text{ m}\] \[x(3) = 3(9) + 5(3) + 2 = 27 + 15 + 2 = 44 \text{ m}\] \[\bar{v} = \frac{x(3) - x(1)}{3 - 1} = \frac{44 - 10}{2} = \boxed{17 \text{ m/s}}\]
Interesting: Here, the average velocity equals the instantaneous velocity at the midpoint (\(t = 2\)). This happens because acceleration is uniform (constant \(a = 6\) m/s²).
Activity: Displacement vs Path Length
L3 Apply
Predict: You walk 4 steps forward and 2 steps backward. Is your displacement equal to, greater than, or less than the total distance walked?
- Stand at a marked starting point. Walk exactly 8 steps forward and mark your position A.
- Now walk exactly 3 steps backward and mark position B.
- Measure the distance from start to A (path 1), then A to B (path 2).
- Calculate: (a) Path length = distance of path 1 + distance of path 2. (b) Displacement = distance from start to B.
- Repeat with different step counts and tabulate.
Observation: If each step is about 0.5 m, then path 1 = 4 m, path 2 = 1.5 m. Path length = 5.5 m. But displacement = 4 − 1.5 = 2.5 m. Displacement is less than path length.
Conclusion: Whenever an object reverses direction, path length exceeds the magnitude of displacement. They are equal only when motion is in one direction without any reversal.
Conclusion: Whenever an object reverses direction, path length exceeds the magnitude of displacement. They are equal only when motion is in one direction without any reversal.
Interactive: Position-Time Graph Explorer L3 Apply
Click anywhere on the graph area to place a point. The slope (velocity) between consecutive points is calculated automatically.
Competency-Based Questions
A delivery van starts from a warehouse (origin O) and travels 5 km east to deliver a parcel at point A. It then continues 3 km further east to point B. After delivering another parcel, it returns 4 km west to point C. The entire journey takes 30 minutes.
Q1. L1 Remember What is the SI unit of displacement?
Answer: B. metre (m). Displacement is a length quantity; its SI unit is the metre.
Q2. L3 Apply Calculate the total path length and the net displacement of the delivery van. (3 marks)
Answer:
Path length = 5 + 3 + 4 = 12 km
Position of C = 5 + 3 − 4 = +4 km (east of O)
Displacement = 4 − 0 = +4 km (east)
Path length = 5 + 3 + 4 = 12 km
Position of C = 5 + 3 − 4 = +4 km (east of O)
Displacement = 4 − 0 = +4 km (east)
Q3. L3 Apply Find the average speed and average velocity of the van for the entire trip. (3 marks)
Answer:
Total time = 30 min = 0.5 h
Average speed = 12/0.5 = 24 km/h
Average velocity = 4/0.5 = 8 km/h (east)
Note: average speed (24) > |average velocity| (8), as expected.
Total time = 30 min = 0.5 h
Average speed = 12/0.5 = 24 km/h
Average velocity = 4/0.5 = 8 km/h (east)
Note: average speed (24) > |average velocity| (8), as expected.
Q4. L4 Analyse The position of a particle varies as \(x = 4t - 2t^2\) (in metres). Determine the time at which the particle returns to the origin and the displacement during this time. (3 marks)
Answer:
At origin, \(x = 0\):
\(4t - 2t^2 = 0 \Rightarrow 2t(2 - t) = 0\)
\(t = 0\) or \(t = 2\) s
The particle is at the origin at \(t = 0\) and returns at \(t = 2\) s.
Displacement from \(t = 0\) to \(t = 2\): \(\Delta x = x(2) - x(0) = 0 - 0 = \boxed{0}\)
The particle returns to the starting point, so net displacement is zero — but the path length is non-zero (the particle first moves forward, then reverses).
At origin, \(x = 0\):
\(4t - 2t^2 = 0 \Rightarrow 2t(2 - t) = 0\)
\(t = 0\) or \(t = 2\) s
The particle is at the origin at \(t = 0\) and returns at \(t = 2\) s.
Displacement from \(t = 0\) to \(t = 2\): \(\Delta x = x(2) - x(0) = 0 - 0 = \boxed{0}\)
The particle returns to the starting point, so net displacement is zero — but the path length is non-zero (the particle first moves forward, then reverses).
Q5. L5 Evaluate A student claims that since the speedometer of a car shows 60 km/h at every instant during a trip, the average speed must be 60 km/h. Is this correct? Can the average velocity also be 60 km/h? Justify. (3 marks)
Answer:
The student is correct about average speed. If the instantaneous speed is constant at 60 km/h throughout the trip, then average speed = 60 km/h.
However, the average velocity equals 60 km/h only if the car moves in a single straight-line direction without turning. If the road curves or the car changes direction, the displacement will be less than the total path length, making |average velocity| < 60 km/h. For example, driving around a circular track at constant 60 km/h gives average speed = 60 km/h but average velocity = 0 after each complete lap.
The student is correct about average speed. If the instantaneous speed is constant at 60 km/h throughout the trip, then average speed = 60 km/h.
However, the average velocity equals 60 km/h only if the car moves in a single straight-line direction without turning. If the road curves or the car changes direction, the displacement will be less than the total path length, making |average velocity| < 60 km/h. For example, driving around a circular track at constant 60 km/h gives average speed = 60 km/h but average velocity = 0 after each complete lap.
Assertion-Reason Questions
Assertion (A): Displacement of an object can be zero even if the distance travelled is non-zero.
Reason (R): Displacement depends only on the initial and final positions, not on the path taken.
Answer: A. If an object returns to its starting position, its displacement is zero (since initial and final positions are the same) regardless of the distance covered. The reason correctly explains this — displacement depends only on endpoints.
Assertion (A): Average speed of an object is always greater than or equal to the magnitude of its average velocity.
Reason (R): Path length is always greater than or equal to the magnitude of displacement.
Answer: A. Since path length ≥ |displacement|, dividing both sides by the same positive time interval \(\Delta t\) gives: path length/\(\Delta t\) ≥ |displacement|/\(\Delta t\), i.e., average speed ≥ |average velocity|. R correctly explains A.
Assertion (A): The instantaneous speed of an object is always equal to the magnitude of its instantaneous velocity.
Reason (R): Instantaneous velocity is defined as the derivative of position with respect to time.
Answer: B. Both statements are true. Instantaneous speed = |instantaneous velocity| is indeed always true. The reason (v = dx/dt) is also correct, but it defines velocity — it does not directly explain why speed equals |velocity|. The explanation lies in the fact that speed is defined as the magnitude of velocity. Hence R is not the correct explanation of A.
Frequently Asked Questions - Position Velocity
What is the main concept covered in Position Velocity?
In NCERT Class 11 Physics Chapter 2 (Motion in a Straight Line), "Position Velocity" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Position Velocity useful in real-life applications?
Real-life applications of Position Velocity from NCERT Class 11 Physics Chapter 2 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Position Velocity?
Key formulas in Position Velocity (NCERT Class 11 Physics Chapter 2 Motion in a Straight Line) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 2?
NCERT Class 11 Physics Chapter 2 (Motion in a Straight Line) is structured so each part builds on the previous one. Position Velocity connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Position Velocity?
CBSE board questions from Position Velocity typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Position Velocity lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
AI Tutor
Physics Class 11 Part I – NCERT (2025-26)
Ready