A particle is moving in the +X direction with its velocity V=ax^1/2 [a is constant] .Assume that at t=0,the particle was located at X=0 then the value of time dependence of velocity

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deepak · Answer

v=ax^1/2 then, dx/dt=v=a(x^1/2) dx/(x^1/2)=a(dt) on intergrating , we get , 2(x^1/2)=at+c since at t=0 => x=0 => c=0 therefore, 2(x^1/2)=at subs, in the first eqn, v=a(at)/2

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A particle is moving in the +X direction with its velocity V=ax^1/2 [a is constant] .Assume that at t=0,the particle was located at X=0 then the value of time dependence of velocity

A particle is moving in the +X direction with its velocity V=ax^1/2 [a is constant] .Assume that at t=0,the particle was located at X=0 then the value of time dependence of velocity

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A particle is moving in the +X direction with its velocity V=ax^1/2 [a is constant] .Assume that at t=0,the particle was located at X=0 then the value of time dependence of velocity

A particle is moving in the +X direction with its velocity V=ax^1/2 [a is constant] .Assume that at t=0,the particle was located at X=0 then the value of time dependence of velocity

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Other Related Questions on Mechanics

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