A point moves in a straight line under the retardation av²,where `a’ is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is:

Question

A point moves in a straight line under the retardation av 2 ,where `a&rsquo; is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is:

Chetan Mandayam Nayakar · Answer

d 2 x/dt 2 =-a(dx/dt)^2, dx/dt= &plusmn;(1/a)d 2 x/dt 2 , distance covered has meaning for positive values only,thus dx/dt=-d 2 x/dt 2 d 2 x/dt 2 +dx/dt=0,characteristic equation is m^2+m=0,m(m+1)=0,x=c1+c 2 e- t , two initial values are required, initial position=c1+c2=0(for the sake of simplicity and without loss of generality),dx/dt=u=-c2, at t=0

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A point moves in a straight line under the retardation av 2 ,where `a’ is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is:

A point moves in a straight line under the retardation av²,where `a’ is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is:

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A point moves in a straight line under the retardation av 2 ,where `a’ is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is:

A point moves in a straight line under the retardation av2,where `a’ is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is:

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A point moves in a straight line under the retardation av²,where `a’ is a positive constant and v is a speed.If the initial velocity is u,the distance covered in t seconds is: