A particle having velocity V = V0 at t = 0 is decelerated at the rate |a| = β √V , where β is a positive constant. Choose the correct option(s) :-(a) The particle comes to rest at t = (2 √V0 ) / β(b) The particle will come to rest at infinty(c) The distance travelled by the particle is (2 V0 3/2 ) / β(d) The distance travelled by the particle is (2 V0 3/2 ) / 3β

a = k(v)^1/2

a = dv/dt = -dv/dt in case of retardation

-dv/dt = k(v)^1/2

 v^-1/2dv = -kdt

 [2v^1/2] lim from u to v    = -k[t]   lim from 0 to t

 2v^1/2 - 2u^1/2 = -kt

 u = v_o (given) so

 2v^1/2 - 2v_o^1/2 = -kt       ..............1

particle comes to rest when final velocity becomes 0 ,

from eq 1 , put v = 0

    t = 2v_o^1/2/k                        (option a)

now use v =  dx/dt

 find a relation in bw x & time ,then put t = 2v_o^1/2 /k u will get the result ...

you have initial velocity u= v0

final velocity, v=0

a= beta/rootV

use v=u+at

you'll get option (a) on isolating t.

simple!!!