## Question

A body is moving towards north with initial velocity of 13m/s. Its is subjected to a retardation of 2 m/s^{2} towards south.The distance travelled in 7th sec is ??

**Joshi sir comment**

displacement of any particular sec. = u+at-(a/2)

here direction of motion is opposite to the direction of acceleration so formula will be u-at+(a/2)

displacement in 7th sec = 0

now we know that direction of motion will become south after 6.5 sec. so distance for last half sec = ut+(1/2)at^{2 }= 1/4

same will be the distance for the first half sec of 7th. so total distance for 7th sec = 1/4 + 1/4 = 1/2

## Read 3 Solution.

using s_{n} = u+a/2(2n-1)

s_{n} is the distance travelled in n^{th} second u=initial vel a=acc^{n}

we get s_{7}=0

hence the ans=0

**IITMAX**10 year ago is this solution helpfull: 5 23

but the answer is not 0 m.

**NIKHIL VARSHNEY**10 year ago is this solution helpfull: 4 14

42

**SAHIL KUMAR**10 year ago is this solution helpfull: 1 13