A PARTICLE HAVING A VELOVITY v=v₀ aT t=0 IS DECELERATED AT THE RATE :|a|=xv^1/2 where x is +ve constantFIND THE DISTANCE TRAVELLED AND TIME WHEN PARICLE COMES TO REST.

Dear vardaan

a=xv^1/2

-dv/dt = xv^1/2

-dv/v^1/2 =xdt

intigrate

 -2v^1/2 =xt +c

given at t=0  ,V=Vo

so -2v_o^1/2 =c

so -2v^1/2 =xt -2v_o^1/2

     v^1/2 =-xt/2 + v_o^1/2  ................1

 when particle come to rest then V=0

       so     0= -xt/2 + v_o^1/2

               t =2/x  v_o^1/2 sec

now again  v =(-xt/2 + v_o^1/2)²

                   ds/dt = (-xt/2 + v_o^1/2)²

                 ds = (-xt/2 + v_o^1/2)² dt

            _o∫^s ds  =_o∫^t (-xt/2 + v_o^1/2)² dt

                S = -2/3x (-xt/2 + v_o^1/2)^{3  limit o to t}

                   = -2/3x (-xt/2 + v_o^1/2)³   +2/3x ( v_o^1/2)³

now put t =2/x  v_o^1/2

                  S =  2/3x  v_o^3/2

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Badiuddin

initial velocity=v, final velocity=v1=0 acc. to the equations of motion v1^2=v^2+2as

where s=distance travelled, a=deceleration of particle

0=v^2-2*xv^1/2*s

v^2=2*xv^1/2*s

v^1/2*v/x*2=s(distance travelled by particle)

2)initial velocity=v, final velocity=v1=0, acc. to equations of motion,

v1=v+at

0=v+at

v=at(as particle is retarding)

v=xv^1/2*t

t=v^1/2/x(time taken)