CBSE Class 9 Answered
A farmer moves along the boundary of square field of side 10m in 40sec . What will be the magnitude of displacement of farmer at the end of 2 min 20 sec from his initial position?
Asked by kanikathakkar836 | 19 May, 2020, 23:27: PM
Let the initial point of the farmer be point P
The distance covered by the farmer in 40 s is 4 × 10 = 40 m.
The total time given: 2 min and 20 s = (2 × 60) + 20 = 140 s
Since the farmer moves 40 m in 40 s, therefore in 1 s the farmer covers a distance,
Therefore, the distance covered by the farmer in 140 s is:
140 × 1 m = 140 m.
Thus the number of rotations to cover 140 m along the boundary is:
So, the farmer will be at point R after 140 s.
So the displacement = PR = = 14.1 m
Answered by Shiwani Sawant | 20 May, 2020, 00:07: AM
Application Videos
Concept Videos
CBSE 9 - Physics
Asked by nittavijayalaxmigayatri | 08 Jun, 2024, 12:36: PM
CBSE 9 - Physics
Asked by sapnamantri05 | 24 Nov, 2023, 16:54: PM
CBSE 9 - Physics
Asked by ashrithpandu84 | 09 Oct, 2023, 20:09: PM
CBSE 9 - Physics
Asked by durgesh21332 | 06 Sep, 2023, 18:17: PM
CBSE 9 - Physics
Asked by leena3732 | 15 Jun, 2023, 09:42: AM
CBSE 9 - Physics
Asked by bablibhati187 | 21 Jun, 2022, 13:48: PM
CBSE 9 - Physics
Asked by npravati227 | 30 May, 2022, 23:04: PM
CBSE 9 - Physics
Asked by chezzlinequeen | 26 May, 2022, 13:03: PM
CBSE 9 - Physics
Asked by urvashi.420 | 10 May, 2022, 17:12: PM