New NCERT Solutions for Chapter 4 Describing Motion Around Us Class 9 Science
- Chapter Name: Describing Motion Around Us
- Textbook Name: Class 9 Science
Chapter 4 Describing Motion Around Us NCERT Solutions Class 9 Science
Page No. 48
Think it Over
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 there is enough time to react and stop the vehicle if the truck suddenly applies the brakes. The exact distance depends on the speed of the vehicle and road conditions.
2. Does this distance depend upon the speed with which we are moving?
Answer
Yes, this distance depends on the speed of the vehicle. At higher speeds, a greater distance should be maintained because the vehicle takes more time and distance to stop safely.
Page No. 51
Pause and Ponder
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?
Answer
The displacement of the athlete will be zero when she returns to her starting point O.
In Fig. 4.4, if the athlete runs from O to A and then comes back to O, her final position becomes the same as her initial position. Therefore, displacement = 0 m.
Total distance travelled = OA + AO = 100 m + 100 m = 200 m.
2. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement
Answer
(i) Fuel consumption depends on the total distance travelled because fuel is used throughout the path covered by the vehicle, irrespective of its direction.
Fuel consumption does not depend on displacement. A vehicle may have zero displacement after returning to its starting point, but fuel is still consumed because distance has been travelled.
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?
Answer
Yes, the ball is moving along a straight inclined track. Therefore, its motion is a straight line motion.
The motion can be represented by a straight line with O as the origin, similar to Fig. 4.3, because the ball moves only in one direction along the track.
Since the ball does not turn back and continues moving in the same direction, the total distance travelled and the magnitude of displacement are equal at all positions A, B, C and D.
Thus,
At A: Distance travelled = Displacement
At B: Distance travelled = Displacement
At C: Distance travelled = Displacement
At D: Distance travelled = Displacement
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
Given:
Distance travelled towards north = 200 km
Distance travelled towards south = 200 km
Total time taken = 3 h + 2 h = 5 h
Average Speed
Total distance travelled = 200 km + 200 km = 400 km
Average speed = Total distance travelled/Total time taken
Average speed = 400/5
Average speed = 80 km h-1
Average Velocity
Displacement = 0 km
(The final position is the same as the initial position.)
Average velocity = Displacement/Total time taken
Average velocity = 0/5 = 0 km h-1
Average velocity = 0 km h-1
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 is equal to the average speed when the object moves in a straight line without changing its direction. In this case, the distance travelled is equal to the magnitude of displacement.
(ii) This occurs when the object returns to its starting point after travelling some distance. In such a case, the displacement is zero, so the average velocity is zero, but the distance travelled is not zero, so the average speed remains non-zero.
Example: A runner completes one full lap of a circular track and returns to the starting point.
Page No. 68
Revise, Reflect, Refine
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
Given:
Distance between home and shop = 250 m
Total Distance Travelled
Home → Shop = 250 m
Shop → Home = 250 m
Home → Shop = 250 m
Shop → Home = 250 m
Total distance travelled = 250 + 250 + 250 + 250
= 1000 m
Total distance travelled = 1000 m
Displacement
The father starts from home and finally returns to home.
Initial position = Final position
Therefore, displacement = 0 m
Displacement from home = 0 m
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
(i) Given:
Height of each floor = 3 m
Ground floor to fourth floor = 4 × 3 m = 12 m
Fourth floor to second floor = 2 × 3 m = 6 m
Total vertical distance travelled = 12 m + 6 m
= 18 m
Total vertical distance travelled = 18 m
(ii) The student starts from the ground floor and finally reaches the second floor.
Displacement = Height of second floor above ground floor
= 2 × 3 m
= 6 m
Displacement = 6 m upward
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 for the scooter to be accelerating even if the speedometer reading remains constant.
The speedometer shows only the speed (magnitude of velocity). Acceleration depends on the change in velocity, and velocity can change due to a change in direction even when speed remains constant.
For example, when the scooter moves along a curved road or takes a turn, its direction changes continuously. Therefore, the scooter is accelerating.
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
Given:
Initial velocity, u = 0 m s-1
Final velocity, v = 24 m s-1
Time, t = 6 s
Average Acceleration
\(a = \frac{v-u}{t}\)
\(a = \frac{24-0}{6}\)
\(a = 4\ \text{m s}^{-2}\)
Average acceleration = 4 m s-2
Distance Travelled
\(s = \frac{u+v}{2}\times t\)
\(s = \frac{0+24}{2}\times 6\)
\(s = 12\times 6\)
\(s = 72\ \text{m}\)
Distance travelled = 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 to come to a stop.
Answer
Given:
Initial velocity, u = 28 m s-1
Final velocity, v = 0 m s-1
Distance travelled, s = 98 m
Using the equation
v2 = u2 + 2as
0 = (28)2 + 2 × a × 98
0 = 784 + 196a
196a = -784
a = -4 m s-2
Acceleration = -4 m s-2
Now, using
v = u + at
0 = 28 + (-4)t
4t = 28
t = 7 s
Time taken to stop = 7 s
6. Fig. 4.27 shows a position-time graph of two objects A and B that are moving along parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
Answer
Yes, objects A and B have equal velocity at one instant of time.
In a position-time graph, the slope of the graph represents velocity.
Object B has a straight-line graph, which means it is moving with constant velocity. Object A has a curved graph, which means its velocity is changing with time.
At the instant where the tangent to the curve of object A becomes parallel to the straight-line graph of object B (near the marked region around 5 s), both graphs have the same slope.
Since equal slopes represent equal velocities, objects A and B have equal velocity at that instant.
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
Both objects start from the same position and reach the same final position at t = 10 s.
Therefore,
Displacement of A = Displacement of B
Time interval = 10 s
Average velocity = Displacement/Time
Hence, both objects have the same average velocity.
Correct Option: (i)
Since both objects move only in one direction and cover the same distance in the same time interval, their average speeds are also equal.
Correct Option: (ii)
Therefore, the correct options are (i) and (ii).
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
Given:
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
Distance travelled, s = ?
Using
\(s = \frac{u+v}{2}\times t\)
\(s = \frac{15+10}{2}\times 36\)
\(s = \frac{25}{2}\times 36\)
\(s = 450\ \text{m}\)
Distance travelled = 450 m
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
First stage: Acceleration
u = 0 m s-1, v = 20 m s-1, t = 5 s
\(s_1 = \frac{u+v}{2}\times t\)
\(s_1 = \frac{0+20}{2}\times 5\)
\(s_1 = 50\ \text{m}\)
Second stage: Constant velocity
v = 20 m s-1, t = 10 s
s2 = vt
s2 = 20 × 10
s2 = 200 m
Third stage: Braking
u = 20 m s-1, v = 0 m s-1, t = 6 s
\(s_3 = \frac{u+v}{2}\times t\)
\(s_3 = \frac{20+0}{2}\times 6\)
\(s_3 = 60\ \text{m}\)
Total distance travelled
= s1 + s2 + s3
= 50 + 200 + 60
= 310 m
Total distance travelled = 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
Given:
Initial velocity, u = 36 km h-1 = 10 m s-1
Reaction time = 0.5 s
Deceleration = 2.5 m s-2
Distance to obstacle = 30 m
Distance travelled during reaction time
s1 = vt
s1 = 10 × 0.5
s1 = 5 m
Braking distance
Using
v2 = u2 + 2as
0 = (10)2 + 2(-2.5)s
0 = 100 - 5s
s = 20 m
Total stopping distance
= 5 + 20
= 25 m
Distance available = 30 m
Stopping distance = 25 m
Since 25 m < 30 m, the bus will stop before reaching the obstacle.
Yes, the bus will be able to stop safely. It will stop 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.
Answer
Rest and motion are relative concepts and depend on the reference point chosen.
An object kept on the Earth can be considered to be at rest with respect to the Earth's surface because its position does not change relative to nearby objects on Earth.
However, with respect to the Sun, the object is not at rest because the Earth itself is moving around the Sun. Therefore, the object is also moving along with the Earth.
Thus, an object on the Earth can be considered at rest with respect to the Earth, but it is in motion with respect to the Sun.
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
(i) The cyclist moves with constant velocity of 3 m s-1 from 20 s to 100 s.
The displacement during this interval is represented by the rectangular area under the velocity-time graph between 20 s and 100 s.
(ii) The velocity decreases from 3 m s-1 to 2 m s-1 during the interval 100 s to 120 s.
The displacement during this interval is represented by the trapezium-shaped area under the graph between 100 s and 120 s.
Displacement: Displacement is equal to the total area under the velocity-time graph.
Area of triangle (0 s to 20 s)
= ½ × base × height
= ½ × 20 × 3
= 30 m
Area of rectangle (20 s to 100 s)
= length × breadth
= 80 × 3
= 240 m
Area of trapezium (100 s to 120 s)
= ½ × (sum of parallel sides) × distance between them
= ½ × (3 + 2) × 20
= 50 m
Total displacement
= 30 + 240 + 50
= 320 m
Displacement = 320 m
Average Acceleration
Initial velocity, u = 0 m s-1
Final velocity, v = 2 m s-1
Time interval, t = 120 s
\(\text{Average acceleration}=\frac{v-u}{t}\)
\(\text{Average acceleration}=\frac{2-0}{120}\)
\(\text{Average acceleration}=\frac{1}{60}\ \text{m s}^{-2}\)
\(\text{Average acceleration}=0.0167\ \text{m s}^{-2}\)
Average acceleration = 0.0167 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 running distance based on the graph.
Answer
The distance travelled is equal to the area under the velocity-time graph.
From the graph, the velocities at different times are approximately:
0 h → 7 km h-1
2 h → 7.5 km h-1
4 h → 7.5 km h-1
6 h → 6.5 km h-1
7 h → 6.5 km h-1
Area from 0 h to 2 h
= ½ × (7 + 7.5) × 2
= 14.5 km
Area from 2 h to 4 h
= 7.5 × 2
= 15 km
Area from 4 h to 6 h
= ½ × (7.5 + 6.5) × 2
= 14 km
Area from 6 h to 7 h
= 6.5 × 1
= 6.5 km
Total distance travelled
= 14.5 + 15 + 14 + 6.5
= 50 km
Running distance ≈ 50 km
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 of 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:
Velocity = 6 m s-1
Time at constant velocity = 2 min = 120 s
Acceleration = 1 m s-2
Time of acceleration = 6 s
Displacement during first 120 s
s1 = vt
= 6 × 120
= 720 m
Final velocity after acceleration
v = u + at
= 6 + (1 × 6)
= 12 m s-1
Displacement during accelerated motion
\(s_2=\frac{u+v}{2}\times t\)
\(s_2=\frac{6+12}{2}\times 6\)
\(s_2=9\times 6\)
\(s_2=54\ \text{m}\)
Total displacement
= 720 + 54
= 774 m
Displacement = 774 m
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.
Answer
For Car A
u = 0 m s-1
v = 5 m s-1
t = 5 s
Acceleration of Car A
\(a=\frac{v-u}{t}\)
\(a=\frac{5}{5}\)
\(a=1\ \text{m s}^{-2}\)
For Car B
u = 0 m s-1
v = 3 m s-1
t = 10 s
Acceleration of Car B
a = (3 - 0)/10
= 0.3 m s-2
Displacement of Car A
= Area under velocity-time graph
= ½ × 5 × 5
= 12.5 m
Displacement of Car A = 12.5 m
Displacement of Car B
= Area under velocity-time graph
= ½ × 10 × 3
= 15 m
Displacement of Car B = 15 m
Points for plotting the graph:
Car A: (0,0), (1,1), (2,2), (3,3), (4,4), (5,5)
Car B: (0,0), (2,0.6), (4,1.2), (6,1.8), (8,2.4), (10,3)
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,
(iv) velocity.
The length of the minute’s hand is 7 cm (Fig. 4.32).
Answer
(i) Length of minute hand = radius = 7 cm
Time interval = 1.5 h = 90 min
The minute hand completes one revolution in 60 min.
In 90 min, it completes 1.5 revolutions.
Distance travelled = 1.5 × Circumference
= 1.5 × 2Ď€r
= 1.5 × 2 × Ď€ × 7
= 21Ď€ cm
= 66 cm (approximately)
Distance travelled ≈ 66 cm
(ii) From 6:00 PM to 7:30 PM, the minute hand moves from the 12 mark to the 6 mark.
The initial and final positions are diametrically opposite.
Displacement = Diameter of the circle
= 2r
= 2 × 7
= 14 cm
Displacement = 14 cm
(iii) Speed = Distance travelled/Time taken
= 66/90
= 0.733 cm min-1
Speed ≈ 0.733 cm min-1
(iv) velocity = Displacement/Time taken
= 14/90
= 0.156 cm min-1
Average velocity ≈ 0.156 cm min-1
