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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.`
7 years ago

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
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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin

```
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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