2. a farmer moves along the boundary of a square field of side 10 m 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?

NCERT Solutions For Class 9 Science Motion IN-TEXT QUESTIONS SOLVED NCERT Textbook Page- 100 Question.1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Answer. Yes, the object inspite of moving through a distance can have zero displacement. Example, if an object travels from point ‘A’ and reaches to the same point ‘A’, then its displacement is ‘O’. Question.2. A farmer moves along the boundary of a square field of side 10 m 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? Answer. Total distance = 40 m Time taken = 40 s Total time taken by the farmer = 2 min 20 seconds =140 seconds … 140 -s- 40 = 3.5 The farmer completes 3 and a half round of the square. i.e., if he starts from A he reaches point ‘C° after 2 minutes 20 seconds. Displacement is AC. Question.3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. Answer. (a) False (b) False NCERT Textbook Page- 102 Question.1. Distinguish between speed and velocity. Answer. Question.2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? – Answer. The magnitude of average velocity of an object is equal to its average speed if the object moves in a straight line in a particular direction. Question.3. What does the odometer of an automobile meas...

The farmer takes 40 s to cover 4 × 10 = 40 m. In 2 min and 20 s (140 s), he will cover a distance = 40/40 x 140 = 140 m Therefore, the farmer completes 140/40 = 3.5 rounds (3 complete rounds and a half round) of the field in 2 min and 20 s. That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point. Now, there can be two extreme cases. Case I: Starting point is a corner point of the field. In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s. Therefore, the displacement will be equal to the diagonal of the field. Hence, the displacement will be√(10 2+ 10 2) = 14.1 m Case II: Starting point is the middle point of any side of the field. In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s. Therefore, the displacement will be equal to the side of the field, i.e., 10 m. For any other starting point, the displacement will be between 14.1 m and 10 m.

Hint: Calculate the net distance travelled by the farmer using the given values and then use it to find the point on the square which is the final position of the farmer. The length of the line joining directly the initial and final position of the farmer is the net displacement. Complete step-by-step answer: Given the problem, a farmer moves along the boundary of a square field of side 10m in 40 seconds. We need to find the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position. First, we calculate the perimeter of the square. The side of square $ABCD$ is $x = 10m$. Perimeter of the square $ = 4 \times side = 4x = 4 \times 10 = 40m$. It is given that the farmer takes 40 seconds to complete one round of this square field. Hence the distance travelled by the farmer in 40 seconds is $40m$. We know that $ = 10\sqrt 2 m \\ $ Hence the net displacement if the farmer after 3.5 rounds is equal to $10\sqrt 2 m$. Therefore, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is $10\sqrt 2 m$. Note: The length of the actual path covered by a body is called the distance covered. The length of the shortest path possible between the initial and final position of the body is called its displacement. Displacement is always less than equal to the distance covered by the body. Visual description in problems like above helps in solving the problem easily.

The correct option is B 14.14 m Given, side of the square field = 10 m Therefore, perimeter = 10 m × 4 = 40 m The farmer moves along the boundary in 40 s. Displacement after 2 m 20 s ( = 2 × 60 s + 20 s = 140 s ) = ? Since in 40 s the farmer moves 40 m, in 1s distance covered by farmer = 40 m 40 = 1 m Therefore, in 140s distance covered = 140 m. Number of rounds covered = 140 m 40 m = 3.5 rounds If the farmer starts from point A, after 3.5 rounds he will at point C of the field. Therefore, displacement A C = √ ( 10 m ) 2 + ( 10 m ) 2 = √ 100 m 2 + 100 m 2 = √ 200 m 2 = 10 √ 2 m = 10 × 1.414 m = 14.14 m