Consider a particle intially moving with a velocity of `5` m `s^(-1)` starts decelerating at a constant rate of `2` m `s^(-2)`.
a. Determine the time at which the particle becomes stationary.
b. Find the distance travelled in the second second.
c. Find the distance travelled in the third second.

Explanation

a. Here `u=5 ms^(-1), a=2 m s^(2), v=0 t=`?
Using `v=u+` at, where
`0=5-2t rArr t=2.5 s`
b. Hnece `u=5 m s^(-1), a=-2 m s^(-2), n=2`.
Using `x_(n)=u+(a)/(2)(2n-1)=5(2)/(2) [(2)-1)]=2 m`
c. Here, if we use the above fromula, we will get `x_(n)=0`. But in reality it is not zero. This fromula is not applicable for the third second becoause velocity becomes zero in the third second, i.e., at `t=2.5 s`. The particle has a turning point at `t=2.5 s`. We have to indirectlu calculate the distance travelled in this distance tramelled between is, we have to derermine the distance travelld between `2g t=t lt=2.5` and `2.5 g t= g t=3`, and then add the two .
Displacemect of the particle at `t=2.5 s` is
`x_(2.5)=(u^(2))/(2a)=(5)^(2)/(2(2))=6.25 m`
Due to sysemtry, the displacement of the pariticle at `t=2 s` and `t=3 s` are same, i.e.,`
Thus, the distance tramelled in the third second is
`x=x_(2.5))-x_(2)+x_(2.5)=x_(3)`
`(6.25-6)+(6.25)-6)=0.5m`.