A particle travelling to East at 2 m/s is uniformly accelerated at 5 m/s^2 for 4 seconds .calculate the displacement of the particle
Answers
Answer:
Ulhas Mandlik, 35, a power-loom owner from Ichalkaranji, Maharashtra, and his mother were homeward-
bound one evening when heavy rain forced them to take shelter beneath a bridge. Not far away, a small
group of labourers huddled together under a part of the cement housing above a 16 metre deep well used
to pump water for irrigation.
Suddenly, Mandlik and his mother heard the labourers scream. When the two got to the well, they were
told that a five year old boy named Hariya had fallen in through a side opening in the structure. Ignoring
his mother's fears, Mandlik quickly knotted together lengths of flimsy rope belonging to the labourers
and asked them to lower him into the dark well. "I hope the rope holds," he thought. As he descended,
Mandlik noticed the metal rungs on the wall of the well. He grabbed hold of one and started climbing
down, when he saw the boy clinging to a pipe running up the well's centre. Grabbing the child, Mandlik
started to climb praying that the old rungs wouldn't give away and plunge them both into the churning
water below. Their luck held and within a few minutes, Mandlik clambered to ground level and handed
over Hariya to his sobbing father.
The man fell at Mandlik's feet and offered him some money as a reward. Refusing the cash, Mandlik took
Hariya and his family to a nearby eatery and offered them steaming tea to warm them up. Several
organisations have honoured Mandlik for his bravery and presence of mind on that wet day three years
ago. "I am happy I was at the right place at the right time," he says," and was able to return a little boy to
his family."
Answer:
Given, initial velocity,
u
=9
i
^
m/s and acceleration,
a
=−2
i
^
m/s
2
Using formula, S
n
=u+
2
1
(2n−1)a, the particle displacement in 5 th sec is
S
5
=9
i
^
+0.5(10−1)(−2
i
^
)=0
Thus, the particle will be reached in the initial position in 5 th sec.
Let, the velocity will be zero at time t.
So using, v=u+at⇒0=9−2t or t=4.5sec
The velocity at 4 sec is v=9−2(4)=1m/s
Using formula, S=ut+(1/2)at
2
, distance covered in time 4 sec to 4.5 sec is S
1
=1(1/2)+(1/2)(−2)(1/2)
2
=0.25m and distance covered in time 4.5 sec to 5 sec is S
2
=(1/2)(2)(1/2)
2
=0.25m (as velocity at 4.5 sec is zero)
Thus, total distance covered in 5 th sec S
1
+S
2
=0.25+0.25=0.5m