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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.

6 years ago

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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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6 years ago

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)

5 years ago

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