yeah true...
a particle has an intial velocity of 9m/s due to east and a const. accerlation of 2m/s^{2}. due to west.
the distance coverd by the particle in the 5th second of itz motion????
a)0
b)0.5m
c)2m
d)none of these

UP 0 DOWN 0 0 10
10 Answers
at time 4, velocity = 98 = 1 m/s
at time 5, velocity = 910 = 1 m/s
velocity=0 at ttime = 4.5
So find the distance between these two intervals 44.5 and 4.55. Add them
there's a direct formula:
dist travelled in tth sec= u + a(2t1)/2
u=initial vel
a=acceleration
t=particular sec
as for here,
d=9 + (2)(2*91)/2
=9  17
= 8
i.e. 8m in west direction.
it is not correct sky...
basically because t=5 not 9
there is one more error.. this is not the formula for the distance travelled but the displacement.
What is happening here is that during the 5th second, for half the time there is +ve displacement and for the other half there is ve. So the displacemnt will become 0.
But distance is not displacemnt. you have to work for individual parts!