NCERT Solutions of Class 9 Science Exploration Chapter 4 “Describing Motion Around Us” explains the fundamental concepts of kinematics, detailing how objects change their position with respect to time in our daily lives. The chapter comprehensively discusses the differences between uniform and non-uniform motion along a straight line. It provides a clear understanding of scalar and vector quantities, explicitly distinguishing between distance vs displacement, speed vs velocity and average acceleration with Revise, Reflect, Refine exercise solutions also.
• इस अध्याय की कुछ बातें हिंदी में हैं (हिंदी मीडियम के छात्रों के लिए)
Chapter 4 Quick Links:
1. Chapter Introduction:
2. Important Points of the Chapter:
3. Common Mistakes Students Make:
4. Intext Questions and Answers:
5. Exercise Questions and Answers:
Chapter Introduction:
The Class 9 Science chapter “Describing Motion Around Us” serves as the ultimate cornerstone for understanding kinematics. It introduces foundational concepts like frame of reference, systematically differentiating between scalar and vector pairs such as distance vs displacement and speed vs velocity. Students explore uniform and non-uniform straight-line trajectories, alongside uniform circular motion. By interpreting complex position-time (s-t) and velocity-time (v-t) graphs, this unit simplifies complex derivations.
Additionally, the chapter emphasizes the graphical representation of motion using distance-time and velocity-time graphs to analyze physical quantities visually. It culminates in the derivation and application of the three core equations of motion (v = u + at, s = ut + ½ at² and v² = u² + 2as) training students to systematically solve practical, real-world numerical problems.
Important Points of the Chapter:
Reference Point (Origin): A fixed point used to specify the exact position, distance and direction of a moving object.
Motion: The state of an object when its position changes with respect to a chosen reference point over time.
Rest: The state of an object when its position relative to a reference point does not change as time passes.
Time Interval: The total duration measured between two distinct clock readings or instants of time.
Distance Travelled: The actual total path length covered by a moving object, specified only by a numerical value without direction.
Displacement: The net change in an object’s position, measured as the shortest straight-line distance from its starting point to its final point in a specific direction.
Magnitude: The numerical value of a physical quantity combined with its standard unit of measurement.
Scalars: Physical quantities that possess only a numerical magnitude and do not require a direction.
Vectors: Physical quantities that require both a specific direction and a numerical magnitude to be completely described.
Average Speed: The total distance travelled by an object divided by the total time interval taken to cover that distance.
Uniform Motion: A type of motion where an object covers equal distances in equal intervals of time along a straight line.
Non-Uniform Motion: A type of motion where an object covers unequal distances in equal intervals of time.
Average Velocity: The total displacement of an object divided by the specific time interval in which that displacement occurs.
Rate of Change: The mathematical ratio representing how much one physical quantity alters in comparison to a corresponding change in time.
Instantaneous Velocity (Velocity): The exact velocity of a moving object measured at a specific single instant of time.
Average Acceleration: The net change in an object’s velocity divided by the total time interval over which that change takes place.
Constant Acceleration: A state of motion where an object’s velocity increases or decreases by exactly equal amounts in equal intervals of time.
Slope: The physical steepness of a plotted graph line that indicates the rate of change of the Y-axis quantity relative to the X-axis quantity.
Kinematic Equations: A set of mathematical formulas that relate displacement, time, initial velocity, final velocity and constant acceleration together.
Uniform Circular Motion: The motion of an object tracing a circular path at a constant, uniform speed while its direction continuously changes.
Common Mistakes Students Make:
Confusing Distance with Displacement: Learner often treat them as the same thing. They forget that distance is the actual total path length covered (a scalar), while displacement is only the shortest straight-line gap between the initial and final positions in a specific direction (a vector).
Assuming Speed and Velocity are Always Equal: A common misconception is that velocity requires specifying a direction along with its magnitude. While average speed depends on total distance, average velocity depends solely on net displacement —meaning a runner complete lap around a track has a high average speed but an average velocity of exactly zero.
Mixing up an ‘Instant of Time’ with a ‘Time Interval’: A slight error happening in use clock readings incorrectly in equations. An instant of time is a single specific point on a clock (like t = 4s), whereas a time interval is the duration or elapsed gap between two distinct instants (like Δ t = t2 – t1).
Misinterpreting the Slope of Different Graphs: A frequent mistake is confusing a position–time (s–t) graph with a velocity–time (v–t) graph. The slope of an s–t graph represents velocity, while the slope of a v–t graph represents acceleration.
Thinking a Flat Horizontal Graph Line Always Means ‘Rest’: A horizontal line does not always indicate that an object is stationary. On a position–time graph, it means the object is at rest. However, on a velocity–time graph, it represents constant velocity with zero acceleration.
Misunderstanding Uniform Circular Motion Acceleration: It is often assumed by students that an object in uniform circular motion travels at a constant, uniform speed, its acceleration must be zero. Recognize the object’s direction is continuously changing along the circular path, its velocity is changing, which means it is actively accelerating.
Forgetting to Check and Convert Units in Numerical: One of the most common errors is solving equations of motion using mixed units (such as plugging in time in minutes or hours directly alongside speed given in meters per second). Always convert values to standard SI units (m, s, ms⁻¹, ms⁻²) before performing calculations.
Applying Kinematic Equations to Variable Acceleration: The three equations of motion (v = u + at, s = ut + ½ at² and v² = u² + 2as) for all numerical problems. They overlook the strict rule that these algebraic formulas are exclusively valid for motion occurring under constant, uniform acceleration.
Intext Questions and Answers:
1. How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
Answer: We should maintain a safe distance from the truck ahead so that our vehicle can stop in time if the truck suddenly brakes. The required distance depends on speed, road conditions, braking efficiency and driver reaction time.
Explanation: We should always keep a safe distance from the truck moving ahead to prevent collisions. When brakes are applied, a vehicle continues moving for some distance before stopping. This stopping distance depends on factors such as the vehicle’s speed, road surface conditions, braking capacity and the driver’s reaction time. If sufficient distance is maintained, the driver gets enough time and space to slow down or stop safely when the truck ahead suddenly applies brakes.
2. Does this distance depend upon the speed with which we are moving?
Answer: Yes, the safe distance depends on our speed. A faster-moving vehicle takes a longer distance to stop after brakes are applied. Therefore, as speed increases, the distance maintained from the vehicle ahead should also increase.
Explanation: Yes, the safe distance between vehicles depends greatly on speed. A vehicle travelling at a higher speed covers more distance before coming to a complete stop after braking. The chapter shows that doubling the speed can greatly increase the stopping distance. Therefore, drivers must maintain a larger gap from the vehicle ahead when moving at higher speeds. Keeping an adequate distance helps avoid accidents and provides enough time for a safe response in emergencies.
Pause and ponder
1. In the example of an athlete running back and forth on a straight track (Fig. 4.4), when 10 cm 20 cm 30 cm will the displacement of the athlete be zero? What will be the total distance travelled in that case?
Answer: The displacement of the athlete will be zero when she returns to the starting point O. In that case, the total distance travelled equals the distance from O to A plus the distance from A back to O, which is 200 m.
Explanation: Displacement becomes zero when the athlete’s final position is the same as her initial position. In Fig. 4.4, this happens if she runs from O to A and then returns to O. Although her displacement is zero because there is no net change in position, she has still covered a distance. The total distance travelled would be OA + AO = 100 m + 100 m = 200 m. This shows that distance and displacement can be different.
2. Fuel used up in a vehicle depends on which of the following? Justify your answer. (i) Total distance travelled (ii) Displacement
Answer: Fuel consumption depends on the total distance travelled, not on displacement. The engine uses fuel throughout the journey according to the distance covered, regardless of the vehicle’s final position relative to its starting point.
Explanation: Fuel used by a vehicle depends on the total distance travelled because fuel is consumed continuously while the vehicle moves. Displacement only tells the net change in position and does not represent the actual path covered. For example, a vehicle may return to its starting point, giving zero displacement, yet it would have consumed fuel throughout the entire journey. Therefore, fuel consumption is related to distance travelled rather than displacement.
3. A ball rolls down an inclined track as shown in Fig. 4.6. Is its motion, a straight line motion? Assuming the starting point of the ball (O) to be the origin, can its O Fig. 4.6: A ball rolling down an inclined track D motion from O to D be depicted using a horizontal line as shown in Fig. 4.3? Are the values of total distance travelled and magnitude of displacement from O equal or different at positions A, B, C and D?
Answer: Yes, the ball’s motion is a straight-line motion along the inclined track. Its motion can be represented on a straight horizontal line. At positions A, B, C and D, the total distance travelled equals the magnitude of displacement.
Explanation: The ball moves along a straight inclined path, so its motion is considered straight-line motion. Taking O as the origin, its motion can be represented on a horizontal line similar to Fig. 4.3 for convenience. Since the ball moves only in one direction without turning back, the total distance travelled from O and the magnitude of displacement are equal at positions A, B, C and D. This follows the rule for motion in one direction along a straight line.
4. During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.
Answer: Total distance travelled = 400 km and total time = 5 h. Therefore, average speed = 80 km/h. Since the final position is the same as the starting position, displacement is zero, so average velocity is 0 km/h.
Explanation: The total distance travelled is 200 km north plus 200 km south, giving 400 km. The total time taken is 3 h + 2 h = 5 h. Therefore,
Average speed = Total distance / Total time
= 400 km / 5 h
= 80 km/h
Since the journey ends at the starting point, the total displacement is zero.
Average velocity = Total displacement / Total time
= 0 km / 5 h
= 0 km/h
Thus, the average speed is 80 km/h, while the average velocity is zero.
5. Under what condition(s) is the (i) magnitude of average velocity of an object equal to its average speed? (ii) magnitude of average velocity of an object zero while its average speed is not zero?
Answer: (i) The magnitude of average velocity equals average speed when the object moves in a straight line without changing direction. In such motion, the distance travelled and the magnitude of displacement are equal.
(ii) This happens when the object returns to its starting position after moving. The displacement becomes zero, making average velocity zero, but the distance travelled remains positive, so average speed is not zero.
Explanation: (i) The magnitude of average velocity is equal to the average speed when an object moves in one direction along a straight line and does not turn back. Under this condition, the total distance travelled is equal to the magnitude of displacement. Since average speed depends on distance and average velocity depends on displacement, both quantities have the same numerical value. Therefore, straight-line motion in a single direction makes average speed and average velocity equal.
(ii) The magnitude of average velocity becomes zero when an object ends its journey at the same position from which it started. In this case, the displacement is zero. However, the object has still travelled some distance during its motion. Since average speed depends on total distance travelled, it remains non-zero. Examples include completing one full lap of a circular track or returning to the starting point after moving along a straight path.
Exercise Questions and Answers
1. My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
Answer: According to question, Total distance travelled = 250 m + 250 m + 250 m + 250 m = 1000 m. As he finally returns to his starting point (home). So, the Displacement from home = 0 m
Explanation: The shop is 250 m away from home.
The father’s journey is:
1. Home to Shop = 250 m
2. Shop to Home = 250 m
3. Home to Shop = 250 m
4. Shop to Home = 250 m
Therefore,
Total distance travelled
= 250 + 250 + 250 + 250
= 1000 m
Since he finally returns to his starting point (home), his initial and final positions are the same.
Displacement
= 0 m
Thus, the father travelled a total distance of 1000 m, while his displacement from home is zero.
2. A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find: (i) the total vertical distance travelled and (ii) their displacement from the starting point.
Answer: To find the distance and displacement, we track the student’s floor changes. Running from the ground to the fourth floor covers 4 x 3 m = 12 m upward. Coming down to the second floor covers 2 x 3 m = 6 m downward. Adding both gives a total distance of 18 m (12 + 6). Since the final position is the second floor, the displacement is 6 m (2 x 3 m) from the ground.
Explanation: Each floor is 3 m above the previous floor.
The student starts from the ground floor.
Ground floor to fourth floor = 4 × 3 m = 12 m upward
Then the student comes down from the fourth floor to the second floor.
Fourth floor to second floor = 2 × 3 m = 6 m downward
Therefore, Total vertical distance travelled = 12 m + 6 m = 18 m
For displacement, compare the initial and final positions:
Initial position = Ground floor (0 m)
Final position = Second floor = 2 × 3 m = 6 m above ground floor
Hence,
Displacement = 6 m upward
So, the student travels a total vertical distance of 18 m and has a displacement of 6 m upward from the starting point.
3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?
Answer: Yes. The scooter can be accelerating even if the speedometer reading is constant. If the scooter is taking a turn, the direction of its velocity changes continuously, causing acceleration.
Explanation: Acceleration occurs whenever velocity changes. Velocity depends on both speed and direction. A speedometer measures only the speed of a vehicle, not its direction.
Therefore, even if the speedometer shows a constant reading, the scooter can still be accelerating if it is moving along a curved path or taking a turn. In such cases, the direction of motion changes continuously, causing a change in velocity and hence acceleration. This is an example of uniform circular motion.
4. A car starts from rest and its velocity reaches 24 m s–1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.
Answer: The car starts from rest and reaches a velocity of 24 m s⁻¹ in 6 seconds. Since the velocity increases uniformly, the average acceleration is obtained from the change in velocity over the given time interval. So, average acceleration is 4 m s⁻². The car covers 72 m while accelerating uniformly from 0 m s⁻¹ to 24 m s⁻¹ in 6 seconds. This means the car travels a total distance of 72 metres in 6 seconds. So, the distance travelled is 72 m.
Explanation: According to the question:
Initial velocity, u = 0 m s⁻¹
Final velocity, v = 24 m s⁻¹
Time, t = 6 s
By Using: a = v-u/t
a = 24-0/6 = 4 m s⁻¹
So, the average acceleration is 4 m s⁻¹.
Now, using the kinetic equation:
s = ut + ½ at²
So, the distance travelled in 6s is 72 m.
5. A motorbike moving with initial velocity 28 m s–1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken (मोटरबाइक का त्वरण और लगा समय) to come to a stop.
Answer: The motorbike is moving with an initial velocity of 28 m s⁻¹ and finally comes to rest after travelling 98 m. Since its velocity decreases uniformly from 28 m s⁻¹ to 0 m s⁻¹, the motorbike experiences a constant negative acceleration. So, the acceleration of the motorbike is –4 m s⁻².
The motorbike slows down uniformly from 28 m s⁻¹ to 0 m s⁻¹ due to the constant negative acceleration. The time required for the motorbike (मोटरबाइक की गति को शून्य तक कम करने में लगने वाला समय) to reduce its velocity to zero is 7 seconds. Hence, the time taken to come to a stop is 7 s.
Explanations: According to question
Initial velocity, u = 28 m s⁻¹
Final velocity, v = 0 m s⁻¹
Displacement, s = 98 m
By Using this formula:
Step 1:
The negative sign shows that the motorbike is slowing down.
Now, using this formula:
Step 2:
Therefore, the acceleration of the motorbike is −4 m s⁻² and the time taken to stop is 7 seconds.
6. Fig. 4.27 shows a position-time graph of two objects A and B ((दो वस्तुएँ A और B जो समानांतर पटरियों पर चल रही हैं) that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
Answer: No, objects A and B never have equal velocity because their position-time graphs have different slopes throughout the motion.
Explanation: The velocity of an object is given by the slope of its position-time graph. In Fig. 4.27, the graph of object A is a straight line with a constant slope, while the graph of object B is steeper. Since the slopes of the two graphs are different at every instant, their velocities are different.
Therefore, objects A and B never have equal velocity, even though they may be moving in the same direction. The steeper graph represents the higher velocity.
7. A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
(i) The average velocity of both over the 10 s time (10 सेकंड के समय में दोनों का औसत वेग) interval is equal since they have the same initial and final positions.
(ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance (समय अंतराल बराबर हैं क्योंकि दोनों समान दूरी तय करते हैं) in equal time.
(iii) The average speed of A over the 10 s time interval is lower than that of B (समय अंतराल B से कम है) since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 10 s time interval is greater than that of B (समय अंतराल B से ज़्यादा है) since B’s speed is lower than A’s in some segments.
Answer: (i) Correct – Both objects have the same displacement in 10 s, so their average velocities are equal. (ii) Correct – Both objects cover the same distance in the same time interval, so their average speeds are equal.
Explanation:
(i) Correct: Average velocity is equal to displacement divided by time. Both objects start from the same position and reach the same final position in 10 seconds. Therefore, their displacements are equal and hence their average velocities are equal.
(ii) Correct: Average speed is equal to total distance travelled divided by time. Both objects move from the same initial position to the same final position in 10 seconds without reversing their direction. Therefore, they cover the same distance in the same time interval and their average speeds are equal.
(iii) Incorrect: This statement is incorrect because object A does not cover a shorter distance than object B. Both objects travel the same distance between the initial and final positions during the 10-second interval.
(iv) Incorrect: This statement is incorrect because average speed depends on the total distance travelled over the entire journey, not on the speed during only some segments of motion.
8. A truck driver driving at the speed of 54 km h⁻¹ notices a road sign (54 km/h⁻¹ की रफ़्तार से गाड़ी चला रहे ट्रक ड्राइवर को सड़क पर एक साइन दिखता है) with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
Answer: To find the distance travelled while slowing down, we apply the average velocity formula (u + v)/ 2 x t. Converting the speeds to m s⁻¹, the initial velocity (u) is 15 m s⁻¹ and the final velocity (v) is 10 m s⁻¹. By Plugging these into the formula, (15+10)/2 x 36 gives the total distance of 450 m.
Explanation: According to question,
Initial velocity, u = 54 km h⁻¹ = 15 m s⁻¹
Final velocity, v = 36 km h⁻¹ = 10 m s⁻¹
Time, t = 36 s
For constant acceleration, Average velocity is:
So, the Distance Travelled
Therefore, the truck travels 450 m while slowing down.
(Note: Using the kinematic equation s = (u+t)/2 x t also give the same result.)
9. A car starts from rest and accelerates uniformly to 20 m s–1 in 5 seconds. It then travels at 20 m s–1 for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
Answer: To Find the Total distance divides the motion into three phases. In stage 1 we applied the formula (u+v/2) x t = we get 50 meter and stage 3 we also same formula for braking (u+v/2) x t = we get 60 meter. And Stage 2 for uniform velocity (v x t) = we get 200 m. Hence, we get 310 meter as a distance.
Explanation: The motion has three stages.
As per the question, Given:
Initial velocity u = 0 m/s
Final velocity v = 20 m/s
Time t = 5s
Stage 1: Acceleration (Distance covered):
S1 = (u + v ) / 2 x t
S1 = (0 + 20 ) / 2 x 5
S1 = 10 x 5 = 50 m
Stage 2: Consistent velocity (Velocity = 20 m/s, Time = 10/s)
S2 = vt
S2 = 20 x 10 = 200 m
Stage 3: Braking (Initial velocity = 20 m/s, final velocity = 0 m/s, time = 6 s)
S3 = (u + v ) / 2 x t
S3 = (20 + 0 ) / 2 x 6
S3 = 10 x 6 = 60 m
Total distance = (50 + 200 + 60) m = 310 m
10. A bus is travelling at 36 km h–1 when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s–2. Will the bus be able to stop before reaching the obstacle?
Answer: To find if the bus stops in time, we calculate the total distance in two steps. Distance during reaction time (S1) as bus travels at 10m s⁻¹ for 5 second, covering 5 m. Now, put braking distance = = v² = u² + 2as, which gives a value of 20 meter. Add both to get total stopping distance as 25 m, which is 5 m ahead of 30 m by seeing the obstacle.
Explanation: Bus Initial speed = 36 km h⁻¹ = 10 m s⁻¹
Reaction time = 0.5 s (driver takes 0.5 seconds to react)
Deceleration = 2.5 m s–2 (bus reduces with constant acceleration)
Distance travelled during reaction time: S1 = vt
Braking distance: v² = u² + 2as
S2= 20 m
Total stopping distance = (5 + 20) m = 25 m
Therefore, 25 m < 30 m, the bus stops 5 m before the obstacle.
11. A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.
Short Answer: Yes, an object on Earth can be considered at rest with respect to the Earth’s surface, but it is not truly at rest when viewed from the Sun because Earth itself is moving.
Detailed Answer: Motion and rest depend on the reference point chosen. An object kept on a table appears to be at rest with respect to the Earth because its position does not change relative to the Earth’s surface.
However, the Earth rotates on its axis and revolves around the Sun. Therefore, when viewed from the Sun or from outer space, the object is also moving along with the Earth. Thus, an object can be at rest relative to one reference frame and in motion relative to another.
12. The velocity-time graph from 0 s to 120 s (0 सेकंड से 120 सेकंड तक वेग-समय ग्राफ़) for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist ((अलग-अलग रंगों में) साइकिल चालक के विस्थापन को दर्शाते हुए) (i) while cyclist is moving with constant velocity (साइकिल चालक एकसमान वेग से चल रहा है) (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration (विस्थापन और औसत त्वरण की गणना करें) in the 120 s time interval.
Answer: The displacement is obtained by adding the areas of the triangle, rectangle and trapezium under the velocity-time graph. Therefore, the cyclist’s total displacement in 120 s is 320 m. The cyclist’s velocity changes from 0 m s⁻¹ to 2 m s⁻¹ in 120 s. Therefore, the average acceleration over the entire interval (पूरे अंतराल के दौरान औसत त्वरण) is 0.017 m s⁻².
Explanation: (Understand the concept of this relation)
(i) Area representing displacement while the cyclist is moving (साइकिल सवार के चलते समय विस्थापन) with constant velocity
The cyclist moves with a constant velocity (स्थिर वेग) of 3 m s⁻¹ from 20 s to 100 s.
So, the displacement is represented by the rectangular area (विस्थापन को आयताकार क्षेत्र द्वारा दर्शाया जाता है) under the graph between 20 s and 100 s (20 सेकंड और 100 सेकंड के बीच ग्राफ़ के नीचे का हिस्सा)।
(ii) Area representing displacement (वेग घटने पर विस्थापन) when the velocity is decreasing
The cyclist’s velocity decreases from 3 m s⁻¹ to 2 m s⁻¹ between 100 s and 120 s.
Hence, the displacement is represented by the trapezium-shaped area under the graph between 100 s and 120 s.
Calculation of Displacement
The total displacement is equal to the total area (कुल विस्थापन कुल क्षेत्रफल के बराबर होता है) under the velocity-time graph.
Area 1: Triangle (0 s to 20 s)
Base = 20 s, Height = 3 m s⁻¹,
Area = ½ × 20 × 3
= 30 m
Area 2: Rectangle (20 s to 100 s)
Length = 100 – 20 = 80 s, Breadth = 3 m s⁻¹
Area = 80 × 3
= 240 m
Area 3: Trapezium (100 s to 120 s)
Parallel sides = 3 m s⁻¹ and 2 m s⁻¹
Distance between them = 20 s
Area = ½ × (3 + 2) × 20
= 50 m
Total Displacement = 30 + 240 + 50 = 320 m
Calculation of Average Acceleration
Initial velocity, u = 0 m s⁻¹,
Final velocity, v = 2 m s⁻¹
Total time, t = 120 s
Average acceleration = (Final velocity – Initial velocity) ÷ Total time
= (2 – 0) ÷ 120
= 1/60
≈ 0.017 m s⁻²
13. A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
Answer: The distance travelled is obtained by estimating the area under the velocity-time graph. The graph can be divided into four phases:
Phase 1 distance ≈ 10.88 km
Phase 2 distance ≈ 11.25 km
Phase 3 distance ≈ 17.50 km
Phase 4 distance ≈ 9.75 km
Adding the distances covered during all four phases: (10.88+11.25+17.50+9.75) ≈ 49.38 km
Explanation:
To estimate the distance travelled, we calculate the area under the velocity-time graph.
Phase 1 (0 h – 1.5 h)
The velocity increases slightly from about 7 km h⁻¹ to 7.5 km h⁻¹.
Average velocity ≈ (7 + 7.5) ÷ 2
= 7.25 km h⁻¹
Distance ≈ 7.25 × 1.5
= 10.88 km
Phase 2 (1.5 h – 3 h)
The velocity remains constant at 7.5 km h⁻¹.
Distance = 7.5 × 1.5
= 11.25 km
Phase 3 (3 h – 5.5 h)
The velocity decreases from 7.5 km h⁻¹ to 6.5 km h⁻¹.
Average velocity ≈ (7.5 + 6.5) ÷ 2
= 7.0 km h⁻¹
Distance ≈ 7.0 × 2.5
= 17.5 km
Phase 4 (5.5 h – 7 h)
The velocity remains constant at 6.5 km h⁻¹.
Distance = 6.5 × 1.5
= 9.75 km
Total Estimated Distance ≈ 10.88 + 11.25 + 17.5 + 9.75
≈ 49.38 km
≈ 49 km
The girl ran approximately 49 km.
14. On entering a state highway, a car continues to move with a constant velocity (एक कार एकसमान वेग से चलती रहती है) of 6 m s–1 for 2 minutes and then accelerates with a constant acceleration (फिर 6 सेकंड के लिए 1 m s⁻² के स्थिर त्वरण के साथ गति बढ़ाता है) 1 m s–2 for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval (2 मिनट 6 सेकंड के समय अंतराल में स्टेट हाईवे पर कार का विस्थापन) by drawing a velocity-time graph for its motion.
Explanation: The motion of the car consists of two parts. First, the car moves with a constant velocity and then it accelerates (एक स्थिर वेग और फिर यह त्वरित होता है) uniformly (समान रूप से)
According to question,
Constant Velocity of a car
Velocity = 6 m s⁻¹
Time = 2 min = 120 s
Distance covered during this interval:
s₁ = v x t
s₁ = 6 x 120
s₁ = 720 m
Thus, the car covers 720 m while moving with constant velocity.
Uniform Acceleration of a car
Initial velocity, u = 6 m s⁻¹
Acceleration, a = 1 m s⁻²
Time, t = 6 s
By Using the equation:
s₂ = (6 x 6) + 1/2 (1) (6)²
s₂ = 36 + 18
s₂ = 54 m
Therefore, the car covers 54 m during (कार एक्सीलरेशन मोशन के दौरान 54 मीटर की दूरी तय करती है) the acceleration motion.
Total displacement = (720 + 54) m = 774 m
15. Two cars A and B start moving with a constant acceleration from rest (विराम अवस्था से एकसमान त्वरण के साथ गति करना), in a straight line. Car A attains a velocity of 5 m s–1 in 5 s. Car B attains a velocity of 3 m s–1 in 10 s. Plot the velocity-time graphs for both (दोनों के लिए वेग-समय ग्राफ़ बनाएँ) the cars in the same graph. Using the graph, calculate the displacement in the two time intervals (दो समय अंतरालों में विस्थापन की गणना करें) mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time (समय के पाँच क्षणों पर उनके वेगों की गणना कीजिए) to plot the graph).
Answers: Both cars start from rest and move with uniform acceleration. The displacement of each car is obtained from the area under its velocity-time graph. Since the graphs are straight lines passing through the origin, the area under each graph forms a triangle. Therefore, the displacement of Car A is 12.5 m (कार A का विस्थापन 12.5 मीटर है) , while the displacement of Car B is 15 m.
Explanation: 1. Car A ka displacement (0 s to 5 s)
2. Car B ka displacement (0 s to 10 s)
(i) Car A (Car A reaches 5 m s⁻¹ in 5 s.)
The displacement is equal to the area under the velocity-time graph.
Displacement = 1/2 x 5 x 5
= 12.5 m (Therefore, the displacement of Car A is 12.5 m.)
(ii) Car B (Car B reaches 3 m s⁻¹ in 10 s.)
The displacement is equal to the area under the velocity-time graph.
Displacement = 1/2 x 10 x 3
= 15 m (Therefore, the displacement of Car B is 15 m.)
Graph Sketch Hints:
The line for Car A will be steeper and will go up to (5,5).
The line for Car B will be less steep and will go up to (10,3).
16. Rohan studies science from 6 PM to 7:30 PM (शाम 6 बजे से 7:30 बजे तक विज्ञान की पढ़ाई करता है) at home. Consider the tip of the minute’s hand (मिनट वाली सुई का सिरा) of the wall clock. During the given time interval, what is its: (i) distance travelled (तय की गई दूरी), (ii) displacement (विस्थापन), (iii) speed and (चाल और) (iv) velocity वेग. The length of the minute’s hand (मिनट की सुई की लंबाई 7 cm है) is 7 cm (Fig. 4.32).
Explanation: Length of the minute’s hand is asRadius of circular path: r = 7 cm
Time interval = 6:00 PM to 7:30 PM = 90 min
The minute hand completes: 90/60 = 1.5 revolutions.
(i) Distance travelled, Circumference of circle: 2 π r
= 2 × 22/7 × 7
= 44 cm
Distance in 1.5 revolutions:
44 × 1.5 = 66 cm
(ii) Displacement, after 1.5 revolutions, the tip reaches the point diametrically opposite its starting position.
Therefore,
Displacement = 2r = 14 cm
(iii) Speed, Total time:
90 min = 5400 s
Speed = 66/ 5400
or, 11/900 = cm/s
(iv) Average Velocity
= 14 / 5400
or, 7/2700 = cm/s
