Guest

A PARTICLE HAVING A VELOVITY v=v 0 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.

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.

Grade:11

2 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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

 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 Experts
Badiuddin

 

 

kedar joshi
35 Points
12 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)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free