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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.
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
o∫s ds =o∫t (-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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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)
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