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A point moves rectilinearly with deceler...

A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the paticles as `alpha=ksqrt(v)`, where k is a positive constant .At the initial movement the velocity of the point in equal to `V_(0)`. What distance will it take to cover that distance?

Text Solution

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Let `t_(0)` be the time in which it comes to a stop.
Given that `-(dv)/(dt)=ksqrt(V)`
`underset(0)overset(t_(0))intkdt=underset(v_(0))overset(0)int-(dv)/(sqrt(v))`
`because kt_(0)=2sqrt(v_(0))`
`therefore t_(0)=(2)/(k)sqrt(v_(0))`
Now to find the distance covered before stopping,
`(dv)/(dt)=(dv)/(ds)(ds)/(dt)=v(dv)/(ds)" But, "(dv)/(dt)=-ksqrt(V),`
`therefore v(dv)/(ds)=-ksqrt(V) therefore sqrt(v)dv=-kds`
`therefore underset(V_(0))overset(0)intsqrt(v)dv=-underset(0)overset(s)intkdsimpliess=(2)/(3k)V_(0)^((3)/(2))`
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