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?
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.
Question 2:
Does this distance depend upon the speed with which we are moving?
Answer
Yes, the distance increases with speed.
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?
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
Question 2:
Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement
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.
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?
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.
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.
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
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?
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).
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.
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.
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).
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).
NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us
