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Particle moves 3m north, then 4m east and finally 6m south. Calculate the...
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Displacement is the shortest distance from the initial to the final position of the particle. The particle moves 3m north, then 4m east and finally 6m south. Refer to the attached image. Here displacesment is AD.
Drop a perpendicular from point A on CD. Let it cuts AD at point E. ABCE is a rectangle.
Side BC= Side AE =4m
Side AB = Side EC = 3m
Therefore DE = CD-EC =6-3=3m
AD^2= AE^2+ED^2 = 4^2+3^2=25
rArr AD = sqrt25 =5m.
The distance travelled is (AB+BC+CD) = 3+4+6 =13m.
Therefore, distance travelled by the particle is 13m and its displacement is 5m.
Posted by llltkl on June 27, 2013 at 4:11 PM (Answer #1)
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