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A particle is moving along a straight line. Its velocity varies as v=6-2t where v in m/s and t in seconds. Find the difference between distance covered and magnitude of displacement in first 4 seconds

A particle is moving along a straight line. Its velocity varies as v=6-2t where v in m/s and t in seconds. Find the difference between distance covered and magnitude of displacement in first 4 seconds
 

Grade:11

1 Answers

IITian
82 Points
2 months ago
V=6-2t implies velocity will be a zero at t=3 sec. And hence maximum displacement 
Also,
V=ds/dot
S=(6-2t)dt  Integrate 
S=6t-t^2
At t=0,6 s=0 
At t=3 sec  ,,s=9metre  and at t=4sec.  S=8m
It will go 9metre and then 1 meter back as at t=4 s=8 which is less than at t=3
Hence displacement =8m and distance =9+1=10 meter 
 
Distance - displacement =10-8=2 meter 
 
Hope it helps 
Do upvote:))

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