NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us
Table of Contents
Page No. 48
Think It Over
Question 1:
How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
टक्कर से बचने के लिए, यदि आगे चल रहा ट्रक अचानक ब्रेक लगा दे, तो हमें उससे कितनी दूरी बनाए रखनी चाहिए?
Answer
Maintain a safe distance of at least 3 seconds behind the truck (more in poor road conditions) to avoid a collision if it suddenly brakes.
यदि आगे चल रहा ट्रक अचानक ब्रेक लगा दे, तो टक्कर से बचने के लिए उससे कम से कम 3 सेकंड की सुरक्षित दूरी बनाए रखनी चाहिए। खराब मौसम या खराब सड़क की स्थिति में यह दूरी और अधिक रखनी चाहिए।
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Question 2:
Does this distance depend upon the speed with which we are moving?
क्या यह दूरी उस गति पर निर्भर करती है जिससे हम चल रहे हैं?
Answer
Yes, the distance increases with speed.
हाँ, यह दूरी वाहन की गति पर निर्भर करती है। जितनी अधिक गति होगी, उतनी ही अधिक सुरक्षित दूरी बनाए रखनी चाहिए।
NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us
Page No. 51
Pause and Ponder
Question 1:
In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?
सीधी पथ (Fig. 4.4) पर आगे-पीछे दौड़ रहे एथलीट का विस्थापन कब शून्य होगा? उस स्थिति में कुल चली गई दूरी कितनी होगी?
Answer
Yes, the displacement is zero because the athlete returns to the starting point.
Total distance
From Fig. 4.4:
O → B = 40 m
B → A = 60 m
A → B = 60 m
B → O = 40 m
Total distance travelled = 40 + 60 + 60 + 40 = 200 m
जब एथलीट दौड़ पूरी करके अपने प्रारम्भिक बिंदु O पर वापस आ जाता है, तब उसका विस्थापन शून्य (0 m) होता है, क्योंकि प्रारम्भिक और अंतिम स्थिति एक ही होती है।
कुल दूरी:
O → B = 40 m
B → A = 60 m
A → B = 60 m
B → O = 40 m
कुल दूरी = 40 + 60 + 60 + 40 = 200 m
अतः, कुल दूरी = 200 m और विस्थापन = 0 m
Question 2:
Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement
वाहन में प्रयुक्त ईंधन निम्नलिखित में से किस पर निर्भर करता है? अपने उत्तर का औचित्य बताइए।
(i) कुल चली गई दूरी
(ii) विस्थापन
Answer
Fuel used in a vehicle depends on (i) the total distance travelled, not on (ii) displacement.
Fuel is consumed throughout the actual path covered by the vehicle. Two journeys can have the same displacement but different distances, leading to different amounts of fuel consumption.
वाहन में प्रयुक्त ईंधन (i) कुल चली गई दूरी पर निर्भर करता है, (ii) विस्थापन पर नहीं।
कारण:
ईंधन वाहन द्वारा तय किए गए वास्तविक मार्ग (कुल दूरी) के दौरान खर्च होता है। दो यात्राओं का विस्थापन समान हो सकता है, लेकिन यदि उनकी कुल दूरी अलग-अलग हो, तो ईंधन की खपत भी अलग होगी। इसलिए ईंधन की खपत कुल दूरी पर निर्भर करती है, न कि विस्थापन पर।
Question 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 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?
चित्र 4.6 में दिखाई गई झुकी हुई पटरियों पर लुढ़कती हुई गेंद की गति क्या सीधी रेखीय गति है? यदि गेंद का प्रारम्भिक बिंदु (O) मूल बिंदु माना जाए, तो क्या O से D तक की गति को चित्र 4.3 की तरह क्षैतिज रेखा पर दर्शाया जा सकता है? क्या A, B, C तथा D बिंदुओं पर कुल दूरी और विस्थापन का परिमाण समान होगा या अलग?
Answer
The motion is not a straight-line motion because the path is curved (inclined track). Total distance and displacement are different at positions A, B, C, and D.
Distance is along the curved path, while displacement is the straight line from O to each point.
नहीं, यह सीधी रेखीय गति नहीं है, क्योंकि गेंद वक्र (मुड़े हुए) मार्ग पर चलती है।
हाँ, यदि O को मूल बिंदु (Origin) माना जाए, तो गेंद की स्थिति को क्षैतिज रेखा पर उसकी प्रगति के अनुसार दर्शाया जा सकता है।
A, B, C तथा D बिंदुओं पर कुल दूरी और विस्थापन का परिमाण समान नहीं होंगे।
(1) कुल दूरी गेंद द्वारा तय किए गए वक्र मार्ग की लंबाई होती है।
(2) विस्थापन O से प्रत्येक बिंदु तक की सीधी रेखा में सबसे कम दूरी होता है।
अतः प्रत्येक बिंदु पर कुल दूरी, विस्थापन के परिमाण से अधिक होगी।
NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us
Page No. 53
Pause and Ponder
Question 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.
एक पारिवारिक सड़क यात्रा के दौरान, आप तीन घंटे में 200 किमी उत्तर दिशा में वाहन चलाते हैं। इसके बाद आप दो घंटे में 200 किमी दक्षिण दिशा में वाहन चलाते हैं। पूरी यात्रा के लिए औसत चाल तथा औसत वेग ज्ञात कीजिए।
Answer
Total distance = 200 + 200 = 400 km
Total time = 3 + 2 = 5 hours
Average speed = \(\frac{400}{5}\) = 80 km/h
Displacement = 0 (since you return to the starting point)
Average velocity = \(\frac{0}{5}\) = 0 km/h
कुल दूरी = 200 + 200 = 400 km
कुल समय = 3 + 2 = 5 hours
औसत चाल = \(\frac{400}{5}\) = 80 km/h
विस्थापन = 0 (क्योंकि आप अपने प्रारम्भिक बिंदु पर वापस आ जाते हैं।)
औसत वेग = \(\frac{0}{5}\) = 0 km/h
Question 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?
किन परिस्थितियों में—
(i) किसी वस्तु के औसत वेग का परिमाण, उसकी औसत चाल के बराबर होता है?
(ii) किसी वस्तु के औसत वेग का परिमाण शून्य होता है, जबकि उसकी औसत चाल शून्य नहीं होती?
Answer
(i) The magnitude of average velocity is equal to average speed when the object moves in a straight line without changing direction.
(ii) The magnitude of average velocity is zero, but average speed is not zero when the object returns to its starting point (displacement is zero, but distance is not zero).
(i) जब कोई वस्तु बिना दिशा बदले सीधी रेखा में गति करती है, तब उसके औसत वेग का परिमाण, औसत चाल के बराबर होता है।
(ii) जब कोई वस्तु चलकर अपने प्रारम्भिक बिंदु पर वापस आ जाती है, तब उसका औसत वेग का परिमाण शून्य होता है, क्योंकि उसका विस्थापन शून्य होता है। लेकिन उसकी औसत चाल शून्य नहीं होती, क्योंकि उसने कुछ दूरी तय की होती है।
NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us
Page No. 68
Revise, Reflect, Refine
Question 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
The father travels between the home and the shop (250 m) four times:
Total distance = 250+250+250+ 250 = 1000m.
Since the starting point is home, and the final position is also home. Thus, displacement is 0 m.
Question 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
Height per floor: 3 m
The student climbs 4 floors.
Distance Up = 4 × 3 = 12 m
Then, he comes down from 4th Floor to 2nd Floor
Distance Up = 2 × 3 = 6 m
(i) Total vertical distance = 12 + 6 = 18 m
(ii) Displacement:
The student’s final position is the 2nd floor, relative to the ground floor.
Displacement (upwards) = 2 floors × 3 m = 6 m
Question 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, it is possible. The speedometer shows only the speed of the scooter, not its direction. If the girl is moving along a curved or circular path at a constant speed, the direction of motion keeps changing. Since velocity changes, the scooter is accelerating even though the speedometer reading remains constant.
Question 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
Initial velocity, u = 0 m/s (starts from rest)
Final velocity, ν = 24 m/s
Time, t = 6 s
Average acceleration:
a = \(\frac{v – u}{t}\) = \(\frac{24 – 0}{6}\) = 4 m/s2
∴ Average acceleration = 4 m/s2
Distance travelled:
S = \(\frac{v^2 – u^2}{2a}\) = \(\frac{24^2 – 0^2}{2 × 4}\)
= \(\frac{576}{8}\)
= 72 m
Question 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 to come to a stop.
Answer
Initial velocity, u = 28 m s-1
Final velocity, v= 0 m s-1 (stops)
Distance, s = 98 m
Acceleration
v2 = u2 + 2as
0 = 282 + 2 × a × 98
0 = 784 + 196a
a = \(\frac{-784}{196}\) = – 4 m/s2
∴ Acceleration = -4 m/s2
(deceleration of 4 m/s2)
Time to Stop
v = u + at
0 = 28 + (-4) × t
t = \(\frac{28}{4}\) = 7 s
∴ Time taken to stop = 7s
Question 6:
Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
Answer
In a position-time graph, the slope (gradient) of the curve at any point gives the instantaneous velocity of the object.
Question 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 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 in equal time.
(iii) The average speed of A over the 10 s time interval is lower than that of B 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 since B’s speed is lower than A’s in some segments.
Answer
From Fig. 4.28, both objects A and B return to the same final position at t = 10 s as their initial position at t = 0 s.
(i) Average velocity of both over 10 s is equal – Correct
Average velocity = \(\frac{Displacement}{Time}\) = \(\frac{0}{10}\) = 0 m/s for both, since their initial and final positions are the same.
(ii) Average speeds are equal.
Question 8:
A truck driver driving at the speed of 54 km h–1 notices a road sign 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
u = 54km h–1 = 54 \(\times\) \(\frac{1000}{3600}\)= 15 m s–1
v = 36 km h–1 = 36 \(\times\) \(\frac{1000}{3600}\) = 10 m s–1
t = 36 s
Now,
Distance travelled, s = \(\frac{u + v}{2} \times t\)
\(\quad\quad\quad\quad\) = \(\frac{15 + 10}{2} \times 36\) = 450 m
Question 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.
Soution
(i) u = 0 ms-1
v= 20 m/s-1
S1 = \(\left(\frac{0 + 20}{2}\right)\times 5\) = 10 \(\times\)5 = 50 m
(ii) S2 = 20 \(\times\)10 = 200 m
(iii) u = 20 ms-1
v= 0 m/s-1
S3 = \(\left(\frac{20 + 0}{2}\right)\times 6\) = 10 \(\times\)6 = 60 m
Total Distance = 50 + 200 + 60 = 310 m
Question 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?
Solution
36 km h-1 = 36 \(\times \frac{1000}{3600}\) = 10 m s-1
(i) Distance covered during reaction time
Reaction Distance, s1 = u x t = 10 m s-1 × 0.5, s = 5 m
(ii) Braking Distance
Given, u = 10 m s-1, v = 0 m s-1
a = -2.5 m/s2
Using,
ν2= u2+ 2as
02= 102+2 (- 2.5)s2
O = 100 – 5s2
5s2 = 100 ⇒ s2 = 20 m
Total Stopping Distance, 5 m + 20 m = 25 m
Since 25 m < 30 m. Yes, the bus stops safely.
Question 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.
Answer
Whether an object is at rest or in motion depends entirely on the frame of reference. Relative to the Earth: An object kept on the Earth (like a chair) is at rest because its position does not change with respect to the Earth’s surface. Relative to the Sun: The same object is in motion because the Earth is rotating on its axis and revolving around the Sun.
Question 12:
The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
(i) while cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.
Answer
Shading the Areas:
1. Constant Velocity: Shade the rectangular area between t =20 s and t = 100 s
2. Decreasing Velocity: Shade the trapezoidal area between t = 1oo s and t = 120 s (where the line slopes downward).
Displacement Calculation (Total Area under the Graph):
Area 1 (Triangle, 0-20s): 12 × 20 × 3 = 30m
Area 2 (Rectangle, 20 – 100s):
length × width = 80 × 3 = 240 m
Area 3 (Trapezium, 100 – 120s):
12 × (3 + 2) × 20 = 50m
Total Displacement: 30+240+50 = 320 m
Average Acceleration = \(\left[\frac{Final Velocity – Initial Velocity}{Total Time}\right]\)
\(\quad\quad\quad\quad\quad\quad\quad\) = \(\frac{2 – 0}{120}\) = \(\frac{1}{60}\) m s-2
Question 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 running distance based on the graph.
Solution
Phase 1 (0 – 1.5 h): Approx. average
velocity is 7.1 km/h.
Distance ≈ 7.1 × 1.5 = 10.65 km
Phase 2 (1.5-3h):
Constant velocity – 7.5 km/h.
Distance 7.5 × 1.5 = 11.25 km
Phase 3 (3 – 5.5h): Velocity drops from
7.5 to 6.5. Average ≈ 7.0 km/h
Distance 7.0 × 2.5 = 17.25 km
Phase 4 (5.5 – Th): Constant velocity is 6.5 km/h.
Distance ≈ 6.5 × 1.5 = 9.75 km
Total Estimated Distance:
10.65 + 11.25 + 17.5 + 9.75 ≈ 49.15 km
Question 14:
On entering a state highway, a car continues to move with a constant velocity of 6 m s–1 for 2 minutes and then accelerates with a constant acceleration 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 by drawing a velocity-time graph for its motion.
Answer
Given :
Constant velocity
v = 6m/s for
2 minutes = 120 s
Then acceleration
a = 1 m/s2for 6s
Displacement during constant velocity (first 120 s)
s1 = ν × t=6 × 120 =720m
Displacement during acceleration (next 6s)
Initial velocity, u = 6 m/s
s2 =ut+ \(\frac{1}{2} \) at2 = 6 × 6+ \(\frac{1}{2} \) × 1 × 62
\(\quad\) = 36 + 18= 54m
Total Displacement s = s1 + s2= 720 + 54 = 774m
Velocity- Time Graph Explanation
From 0 to 120 s → horizontal line at 6 m/s (constant velocity)
From 120 to 126 s → straight slopping line uniform acceleration from 6 m/s to 12 m/s
Displacement = area under the graph (rectangle + trapezium)
Question 15:
Two cars A and B start moving with a constant acceleration from rest 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 the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals
(Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).
Answer
Car A : a = \(\left(\frac{v-u}{t}\right)=\left(\frac{5-0}{5}\right) \)= 1 m/s2
Car B : a = \(\left(\frac{3-0}{10}\right)\) = 0.3 m/s2
Displacements:
Car A (in 5s) : s = \(\frac{1}{2}\) × 5 × 5= 12.5 m
Car B (in 10s) : s = \(\frac{1}{2}\) × 10 × 3 = 15 m
Question 16:
Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.
The length of the minute’s hand is 7 cm (Fig. 4.32).
Answer
Length of minute hand (r): 7cm
Time interval: 6:00 PM to 7:30 PM = 90 mins = 5400 s
Rotations: In 90 minutes, the hand completes 1.5 circles.
(i) Distance is the total (circumference) covered
Distance = 1.5 × (2πr)
Distance = 1.5 × 2 × \(\frac{22}{7}\) × 7
\(\quad\quad\)= 1.5 × 44 = 66 cm
(ii) Displacement is the straight-line distance from the start (12 O’clock) to the end (6 0’ clock)
This is equal to the diameter of the clock
Displacement = 2 × 7 = 14 cm
(iii) Speed = \(\frac{\text { Distance }}{\text { Time }} \)
Speed = \(\frac{66 \mathrm{~cm}}{90 \mathrm{~min}} \) = 0.733 cm/min
in cm/s: \( \frac{66}{5400}=\frac{11}{900}\) = cm/s
(iv) Velocity = \(\frac{\text { Distance }}{\text { Time }}\)
Velocity= \(\frac{14 \mathrm{~cm}}{90 \mathrm{~min}} \) = 0.155cm/min
\(\quad\)= \(\frac{14}{5400}=\frac{7}{2700}\) cm/s
NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us
