# motion in 1-D

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

62
Lokesh Verma ·

at 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

1
skygirl ·

there'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.

62
Lokesh Verma ·

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!

1
skygirl ·

ok!
but i think its not clear :(

62
Lokesh Verma ·

hmm.. i will post an image to make it clear!

62
Lokesh Verma ·

1
skygirl ·

THANKYOU!!!!!!!
PERFECTLY CLEAR!!!
THANX ONCE AGAIN!!!

62
Lokesh Verma ·

a picture is worth a thousand words

1
skygirl ·

yeah true...

1
mydreamsbig ·

thnx nishant