Question

A particle moving with constant acceleration of 2m/s² due west has an initial velocity of 9m/s due east.find the distance covered in fifth second of its motion??

Answer 1

hi frnd

initial velocity u=+9m/s ;


acceleration a =-2m/s²

in this problem,acceleration's direction is opposite to the velocity's direction

let 't' be the time taken by the particle to, reach a point where it makes a turn along the straight line.

we have,v=u+at

0=9+2t

we get,t=4.5s

now let us find the distance covered in 1/2 second i.e..,from 4.5 to 5 second

let u=0at t=4.5 second

then distance covered in 1/2 s

s=1/2at²

which is equal to 1/4

total distance covered in fifth second of its motion is given by s=2s=2(1/4)=0.5m


I believe this will help u :)

Answer 2

Here, basically we should use vector rule for kinematics equation of motion.
acceleration = 2m/s² in West direction
in vector form ,
a = -2i m/s²

initial velocity = 9 m/s² in East direction ,
in vector form ,
u = 9i m/s

now , use kinematic equation ,
Snth = u + a(2n -1)/2

here Snth , u , and a all are in vector form Snth is displacement in nth second
u is intial velocity of particle
a is acceleration of particle .

now,
Snth = 9i + (-2i)(2 × 5 - 1)/2
= 9i +(-2i)(9)/2
= 9i - 18i/2
=9i - 9i
= 0

hence, particle displacement in 5th second = 0 .
it means particle reached 5th second in intial position .

===================
for distance
———————
here we see displacement = 0 in 5th second . it means in 5th second final position = initial position.

first we find out where , velocity becomes zero
use ,
V = u + at
0 = 9 -2t
t = 4.5 sec

now , find out velocity at 4 sec
v = u + at
v = 9 - 2 × 4 = 1m/s

so, distance covered in 4sec to 4.5 sec
= ut + 1/2at²
= 1 × 1/2 - 1/2 × 2 × 1/4
= 0.25 m

again,
distance covered in 4.5 sec to 5 sec ,
velocity at 4.5 sec = 0
so,
Distance = 1/2at²
=1/2 × 2 × 1/4
= 0.25 m

hence, distance covered in 5th second = distance covered in 4 sec to 4.5 sec + distance covered in 4.5 sec to 5sec
= 0.25m + 0.25m
= 0.5 m