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vardaan kumar Grade: 11
        

A PARTICLE HAVING A VELOVITY v=v0 aT t=0 IS DECELERATED  AT THE RATE :|a|=xv1/2 where x is +ve constantFIND THE DISTANCE TRAVELLED AND TIME WHEN PARICLE COMES TO REST.

7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear vardaan


a=xv1/2


-dv/dt = xv1/2


-dv/v1/2 =xdt


intigrate


 -2v1/2 =xt +c


given at t=0  ,V=Vo


so -2vo1/2 =c


so -2v1/2 =xt -2vo1/2


     v1/2 =-xt/2 + vo1/2  ................1


 when particle come to rest then V=0


       so     0= -xt/2 + vo1/2


               t =2/x  vo1/2 sec


 


now again  v =(-xt/2 + vo1/2)2


                   ds/dt = (-xt/2 + vo1/2)2


                 ds = (-xt/2 + vo1/2)2 dt


            os ds  =ot (-xt/2 + vo1/2)2 dt


                S = -2/3x (-xt/2 + vo1/2)3  limit o to t


                   = -2/3x (-xt/2 + vo1/2)3   +2/3x ( vo1/2)3


now put t =2/x  vo1/2


                  S =  2/3x  vo3/2


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7 years ago
kedar joshi
35 Points
										

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)

6 years ago
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