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Referrring to the v^(2)-s diagram of a p...

Referrring to the `v^(2)-s` diagram of a particle, find the displacement of the particle durticle during the last two seconds.
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Text Solution

Verified by Experts

Slope of `v^(2)-s` graph`
`m=-((2000-1500))/(100-50)=-10`
Relation between `v_(2)` and s: `v^(2)=-10 s+C`
At `s=50, v^(2)=2000`
`rArr 2000=-10xx50+C rArr C=2500`
`v=-10 s+2500`
`rArr (2v dt)/(dt)=-10(ds)/(dt)`
`rArr 2va=-10 v rArr a=-5 ms^(-2) rarr` constant
To find initial velocity:
Put `s=0, v=u`
`rArr u^(2)=2500 rArr u=50 ms^(-1)`
Apply `v=u+at`
`rArr 0=50-5 T`
`rArrT=10 s rarr` time of motion
`Displacement during last two seconds :
`S=S_(t)=10s-S_(t)=8s`
`=[50 xx 10 -(1)/(2)5 (10)^(5)]-[50-(1)/(2)5(8)^(2)]=10m`.
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