NCERT Class 9 Science Chapter 4 Describing Motion Around Us Solutions English Medium As Per New Syllabus. NCERT Class 9 Science Chapter 4 Describing Motion Around Us Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 9 Science Chapter 4 Describing Motion Around Us Question Answer Download PDF and select needs one. CBSE Class 9 Science Solutions.
NCERT Class 9 Science Chapter 4 Describing Motion Around Us
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 9 Science Textbook Solutions. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 9 Science Notes in English Medium Solutions for All Subject, You can practice these here.
Describing Motion Around Us
Chapter – 4
| 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?
Ans: The displacement of the athlete will be zero when the athlete returns to the starting point. In that case, the total distance travelled will be the distance covered in going and coming back (twice the one-way distance).
2. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled.
(ii) Displacement.
Ans: Fuel used depends on (i) Total distance travelled because fuel is consumed throughout the journey, irrespective of the 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 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?
Ans: The ball moves along a straight inclined track, so its motion is a straight-line motion. Yes, its motion from O to D can be represented by a horizontal line showing the positions. At positions A, B, C, and D, the total distance travelled and the magnitude of displacement are equal because the ball moves in one direction without changing its path.
| Pause And Ponder |
1. 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.
Ans: Total distance travelled = 200 km + 200 km = 400 km
Total time = 3 h + 2 h = 5 h
Average speed = 400 ÷ 5 = 80 km/h
The final position is the same as the starting position, so displacement = 0 km.
Average velocity = 0 ÷ 5 = 0 km/h
2. Under what condition(s) is the:
(i) magnitude of average velocity of an object equal to its average speed?
Ans: When the object moves in a straight line in one direction without changing direction.
(ii) magnitude of average velocity of an object zero while its average speed is not zero?
Ans: When the object returns to its starting point, making the displacement zero while the distance travelled is not zero.
| Pause And Ponder |
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?
Ans: Total distance travelled: The path consists of four trips:
Home to shop (250 m) + Shop to home (250 m) + Home to shop (250 m) + Shop to home (250m).
Total Distance = 250 + 250 + 250 + 250 = 1000 m.
Displacement: Displacement is the net change in position between the starting and ending points.
Since he started at home and finally returned home, his net change in position is 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.
Ans: Total vertical distance travelled:
The student goes up 4 floors (4 × 3 m = 12 m) and then down 2 floors from the 4th floor to the 2nd floor (2 × 3 m = 6 m)
Total distance = 12 m + 6 m = 18 m
(ii) Their displacement from the starting point.
Ans: Displacement: The starting point is the ground floor (0 m) and the ending point is the 2nd floor (2 × 3 m = 6 m) Displacement is the shortest straight-line distance, which is 6 m upwards
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?
Ans: Yes, it is possible. The speedometer only shows the magnitude of velocity (speed). Acceleration occurs when velocity changes, and velocity has both a magnitude and a direction. If she is riding on a curved path or turning a corner at a constant speed, the direction of her velocity is continuously changing, which means she 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.
Ans: Given:
Initial velocity, u=0m/s
Final velocity, v=24m/s
Time, t=6s
Average Acceleration
Using the formula:
a = (v – u) / t
a = (24 – 0) / 6
= 24/6
= 4 m/s2
So, the average acceleration is 4 m s⁻²
Distance Travelled
Using the equation:
s = (0× 6) + 1/2 (4) (6)2
s = 0 + 2 × 36
s = 72m
So, the distance travelled in 6 s 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 to come to a stop.
Ans: Given:
Initial velocity, u=28m/s
Final velocity, v=0m/s
Distance travelled, s=98m
Acceleration
Using the equation:
V2 = u2 + 2as
0 = (28)2 + 2 (a) (98)
0 = 784 + 196 a
196a = – 784
a = -784/196
= – 4m/s2
So, the acceleration of the motorbike is –4 m s⁻²
Time Taken to Stop
Using the equation:
V = u + at
0 = 28 + (-4)t
4t = 28
t = 7s
So, the time taken to come to a stop is 7 s.
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.
Ans: Yes, their velocities are always equal. In a position-time graph, the velocity is determined by the slope (steepness) of the line. Because lines A and B are strictly parallel, their slopes are identical, meaning they have the exact same constant velocity throughout their motion.
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.
Ans: (i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.
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.
Ans: Given:
Initial speed, u=54km h−1
Final speed, v=36km h−1
Time taken, t=36s
Step 1: Convert speeds into m/s
Step 2: Find acceleration
Using the formula:
a = (v – u) / t
a = (10 – 15) / 36
a = -5/36 m/s2
Step 3: Find the distance travelled
Using the equation:
s = ut + ½ at2
s = (15 × 36) + 1/2 (-5/36) (36)2
s = 540 – 90
s = 450 m
The truck travelled 450 m during this time.
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.
Ans: Given:
u=0, v=20m/s, t=5 s
Constant speed 20m/s for 10s
Stops in 6 s
Part 1: Accelerating
s = ut + 1/2 at2
a = (20 – 0) / 5
= 4 m/s2
s1 = 0 + 1/2 (4) (5)2 = 50 m
Part 2: Constant Speed
S2 = vt = 20 × 10 = 200m
Part 3: Braking:
a = (0 – 20) / 6 = -10/3 m/s2
s = ut + ½ at2
s = 20 (6) + ½ (-10/3) (6)2 = 60 m
Total Distance
s= 50 + 200 + 60 = 310m
The total distance travelled is 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?
Ans: Given:
Speed of bus =36km/h=10m/s
Obstacle distance =30m
Reaction time =0.5s
Retardation =2.5m/s2
Reaction Distance
Distance=vt=10×0.5=5m
Braking Distance
Using:
V2 = u2 + 2as
0=(10)2 +2(−2.5)s
0=100−5s
s=20m
Total Stopping Distance
=5+20=25m
Since the total stopping distance is 25 m, which is less than the obstacle distance (30 m), the bus will stop safely before reaching 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.
Ans: Whether an object is considered at “rest” depends entirely on the chosen reference point. If the reference point is a fixed location on Earth, the object is at rest. However, if the reference point is the Sun or an observer in space, the object is in continuous motion along with the Earth.
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, and.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.
Ans: (i) The displacement when the cyclist is moving with constant velocity is represented by the rectangular area under the graph from 20 s to 100 s.
(ii) The displacement when the velocity of the cyclist is decreasing is represented by the trapezium-shaped area under the graph from 100 s to 120 s.
Calculation of Displacement:
Displacement = Area under the velocity-time graph
(a) Area of triangle (0 s to 20 s)
= 1/2 × 20 × 3 = 30 m
(b) Area of rectangle (20 s to 100 s)
= 80 × 3 = 240 m
(c) Area of trapezium (100 s to 120 s)
= 1/2 (3 + 2) × 20 = 50 m
Total Displacement
= 30 + 240 + 50 = 320m
Calculation of Average Acceleration
Average acceleration = (Final Velocity − Initial Velocity) / Total Time
a = (2 − 0)/120 = 1/60 = 0.0167 m/s²
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.
Ans: The area under the velocity-time graph gives the distance.
Based on the axes in Fig. 4.31 (Time in hours, Velocity in km/h):
Area 1 (Triangle from 0 to 2 h): ½ ×base (2 h) × height (5 km/h) = 5 km.
Area 2 (Rectangle from 2 to 4 h): base (2 h)×height (5 km/h)= 10 km.
Area 3 (Triangle from 4 to 6 h): ½ × base (2 h)×height (5 km/h) = 5 km.
Total distance: 5+10+5= 20 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 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.
Ans: Part 1 (Constant Velocity):
u = 6 m/s
t = 2 min = 120 s
s₁ = ut = 6 × 120 = 720 m
Part 2 (Acceleration): u=6 m/s, a=1 m s⁻², t=6 s.
Displacement s2 =ut+ ½ at2 =(6×6)+ ½ (1)(6)2 = 36 + 18= 54 m.
Total Displacement: s1 +s2 =720 + 54= 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 in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).
Ans: For Car A,
u = 0,
v = 5 m/s
t = 5s
a = (v-u)/t
= (5-0)/5
= 1 m/s²
For Car B,
u = 0,
v = 3 m/s
t = 10s
a = (v- u)/t
= (3 – 0) / 10
= 0.3 m/s2
Displacement of Car A
Displacement = Area of triangle OAC
= 1/2 × 5 × 5
= 12.5 m
Displacement of Car B
Displacement = Area of triangle OBD
= 1/2 × 10 × 3
= 15 m
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).
Ans: Time = 6:00 PM to 7:30 PM = 90 min
Length of minute hand (radius), r=7 cm
In 90 minutes, the minute hand completes:
90/60 = 3/2
(i) Distance travelled:
Distance travelled by the tip of the minute hand:
Distance travelled = (3/2) × 2πr
= (3/2) × 2 × (22/7) × 7
= 66 cm
(ii) Displacement:
At 6:00 PM, the minute hand is at 12.
At 7:30 PM, the minute hand is at 6.
The tip moves from one end of the diameter to the other.
Displacement = 2r
= 2 × 7
= 14 cm
(iii) Speed
Speed = Distance / Time
= 66/90
= 0.733 cm/min
(iv) Velocity
Velocity = Displacement / Time
= 14 / 90
= 0.156 cm/min
Subject Ka Complete Video Solution Dekhein
Step-by-step easy concept explanation aur full questions Answer ke liye watch karein ya app download karein:
Buy Now