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NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science Chapter 8 Motion are prepared by CBSE student eCARE expert to score good marks in class 9. This chapter contain many topics which are very important to score good in class 9. There are some important topic/Activity mentioned below- 

Topics and Sub Topics in Class 9 Science Chapter 8 Motion:

  1. Motion
  2. Describing Motion
  3. Measuring the Rate of Motion
  4. Rate of Change of Velocity
  5. Graphical Representation of Motion
  6. Equations of Motion by Graphical Method
  7. Uniform Circular Motion



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Some questions are asked by class 9 students related to Science

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Solution

Yes, an object moving a certain distance can have zero total displacement. Displacement refers to the shortest distance between the initial and the final positions of the object. Even if an object moves through a considerable distance, if it eventually comes back to its initial position, the corresponding displacement of the object would be zero.

2. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Solution

Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance traveled by the farmer in 40 seconds is 4*(10) = 40 meters.

Therefore, the average distance covered by the farmer in one second is: 40m/40 = 1m

Two minutes and 20 seconds can be written as 140 seconds. The total distance traveled by the farmer in this timeframe is: 1 m * 140 = 140m

Since the farmer is moving along the boundary of the square field, the total number of laps completed by the farmer will be: 140m/40 = 3.5 laps

Now, the total displacement of the farmer depends on the initial position. If the initial position of the farmer is at one corner of the field, the terminal position would be at the opposite corner (since the field is square).

In this case, the total displacement of the farmer will be equal to the length of the diagonal line across the opposite corners of the square.

Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows: √(102+102)√20014.14m.

This is the maximum possible displacement of the farmer.

If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10m. This is the minimum displacement.

If the farmer starts at a random point around the perimeter of the square, his net displacement after traveling 140m will lie between 10m and 14.14m.