NCERT Solutions for Science Class 9 Chapter 7

Chapter 7: Motion

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer: Yes. An object can have zero displacement if it has moved through a distance.
Displacement is defined as the shortest distance from the initial point to the final point.
Hence, if the starting (initial) point is the same as the final point then the displacement of the object is zero.
Suppose a man is walking in a square park of length 20m. He starts from point A and walks along all the corners of the park through points B, C and D and comes back to the same point A.
The total distance covered by the man \(20m+20m+20m+20m=80m\).
As the starting point and final point are same, the shortest distance between his initial and final position is zero
Therefore, the displacement is zero.
Question 2: A farmer moves along the boundary of a square field of side 10m in 40s .What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer: It is given that,

Farmer takes 40 s to cover a square field of side 10 m.

\(⇒Distance=4\times10=40\space m\)

It is known that, \(Spee=\frac{Distance}{Time}\)

\(⇒Speed=\frac{40}{40}=1\)

Therefore, speed of the farmer is \(1\space m/s\).

In 2 minutes 20 seconds distance travelled is \(Speed\times Time\).

\(⇒Distance=1\times(2\times60+20)\)

\(⇒Disatance =140\space m\)

Number of rounds farmer covered \(=\frac{140}{40}=3.5\)

After 2 minutes 20 seconds the farmer will be at the opposite end of starting point, completing 3 and half rounds.

1. If the farmer starts from any corner of the field: The displacement will be equal to the diagonal of the field.

\(⇒Displacement=\sqrt{10^2+10^2}=14.14\space m\)

Question 2: If the farmer starts from the middle point of any side of the field: The final point will be the middle point of the side opposite to the initial point.

\(⇒Displacement=10\space m\)

Therefore, the magnitude of displacement if the farmer starts at any corner is 14.14 m and if the farmer statrs from middle point of any side is 10 m.

Question 3: Which of the following is true for displacement?
1. It cannot be zero.
Answer: Not true. When the initial and final position of the object is the same, then the displacement is zero.

2. Its magnitude is greater than the distance travelled by the object.
Answer: Not true. Displacement is the measure of the shortest distance between the initial and final position of an object.
Therefore, it is always smaller than or equal to the magnitude of distance travelled by the object.

Question 4: An artificial satellite is moving in a circular orbit of radius 42250km. Calculate its speed if it takes 24 hours to revolve around the earth?
Answer: It is given that,
Radius of the circular orbit,

\(r=42250\space km\)

Time taken by the satellite to revolve around earth, \(t=24\space h\)

Speed of the artificial satellite, \(v=?\)

It is known that,

\(v=\frac{2\pi r}{t}\)

\(⇒v=\frac{2\times3.14\times42250}{24}\)

\(⇒v=1.105\times10^4\space km/h\)

\(⇒v=\frac{1.105\times10^4}{3600}\space km/s\)\

\(⇒v=3.069\space km/s\)

Therefore, the speed of the artificial satellite is \(v=3.069\space km/s\).

Question: Distinguish between speed and velocity.
Answer: The differences between speed and velocity are as follows:

Question 6: Under what condition(s) is the magnitude of average velocity of an object added equal to its average speed?

Answer: It is known that

Average speed \(=\frac{Total\space distance\space covered}{Total\space time\space taken}\)

Therefore, the magnitude of average velocity of an object is equal to its average speed when total distance covered is equal to the displacement.
Question 7: What does the odometer of an automobile measure?
Answer: The distance covered by an automobile is recorded by the odometer of an automobile.

Question 8: What does the path of an object look like when it is in uniform motion?
Answer: An object has a straight-line path when it is in uniform motion.
Question 9: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, \(3\times10^8ms^{-1}\)

Answer: It is given that,

Time taken by a signal to reach ground from a spaceship \(=5\space min=5\times60=300\space sec\)

Speed of the signal is equal to speed of light

\(=3\times10^8ms^{-1}\)

It is known that,

\(Speed=\frac{Distance\space travelled}{Time\space taken}\)

\(⇒Distance\space travelled=Speed\times Time\space taken\)

\(⇒Distance\space travelled=3\times10^8\times300=9\times10^{10}m\)

Therefore, the distance of the spaceship from the ground station is \(9\times10^{10}m\).

Question 10: When will you say a body is in
1. uniform acceleration?
Answer: When the magnitude and the direction of acceleration of a body is constant i.e., velocity changes at an equal rate then the body is said to be in uniform acceleration.

2. non-uniform acceleration?
Answer: When the acceleration of a body changes in magnitude or direction or both i.e., velocity changes at an unequal rate then the body is said to be in non-uniform
acceleration.

Question 11: A bus decreases its speed from \(80\space kmh^{-1}\) to \(60\space kmh^{-1}\) in 5s. Find the acceleration of the bus.

Answer: It is given that,

Intial speed of the bus, \(u=80\space km/h\)

\(⇒u=80\times\frac{5}{18}m/s=22.22\space m/s\)

Final speed of the bus, \(v=60\space km/h\)

\(⇒v=60\times\frac{5}{18}m/s=16.66\space m/s\)

Time taken to decrease speed, \(t=5s\)

It is known that,

Acceleration, \(a=\frac{v-u}{t}\)

\(⇒a=\frac{16.66-22.22}{5}\)

\(⇒a=-1.112\space m/s^2\)

Therefore, the acceleration of the bus is \(-1.112\space m/s^2\). The negative sign indicates that the velocity of the car is decreasing. Decreasing acceleration is called retardation.

Question 12: A train starting from a railway station and moving with uniform accelerationattains a speed 40 km/h in hin 10 minutes. Find its acceleration.

Answer: It is given that,

Intial velocity of the train, \(u=0\)  (The train is starting from rest)

Final velocity of the train, \(v=40\space km/h\)

\(⇒v=40\times\frac{5}{18}mm/s=11.11\space m/s\)

Time taken, \(t=10\times60=600\space s\)

It is known that,

Acceleration, \(a=\frac{v-u}{t}\)

\(⇒a=\frac{11.11-0}{600}\)

\(⇒a=0.0185\space m/s^2\)

Therefore, the acceleration of the train is \(0.0185\space m/s^2\)

Question 13: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer: The distance-time graph for uniform motion of an object is a straight line.