1there's a direct formula:
dist travelled in t-th sec= u + a(2t-1)/2
u=initial vel
a=acceleration
t=particular sec
as for here,
d=9 + (-2)(2*9-1)/2
=9 - 17
= -8
i.e. 8m in west direction.
a particle has an intial velocity of 9m/s due to east and a const. accerlation of 2m/s2. 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
62at time 4, velocity = 9-8 = 1 m/s
at time 5, velocity = 9-10 = -1 m/s
velocity=0 at ttime = 4.5
So find the distance between these two intervals 4-4.5 and 4.5-5. Add them
162it 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!
