Question

A particle has an initial velocity of 9 metre per second due east at a constant acceleration of 2 metre per second square due West the distance covered by the particle in the third second of its motion is


tell fast please​

Answer 1

HERE IS YOUR ANSWER -----

Here the particle velocity will become zero in 4.5 seconds, as

V=U+at

0=9–2*t

t=4.5sec.

Displacement in the same period will be

v^2-U^2=2*a*S

-81=-2*S

S4.5=20.25m

Let us call the point of velocity being zero as Point A.

now body will start moving backwards,

displacement in next 0.5 sec from point A,

s=U*t+1/2*a*t^2

S0.5=1/2*(-2)*0.25= -0.25m

hence distance covered is 0.25 m

PLEASE PLEASE MARK ME AS A BRAINLIEST

Answer 2

hence distance covered ke badd 2 image click