sir,

how to calculate the distance travelled by a body in 1d motion or freely falling motion during the last second of its motion..

and 1 question is there in 1d motion

u=9m/s due east, a=2 m/s2 due west.

find the distance travelled by the body during its 5th second of motion

dear dashmesh , here is the simple method to solve such type of problems ....

distance travelled in n seconds (S_n) = gn²/2

distance travelled in (n-1) seconds (S_n-1) = g(n-1)²/2

distance travelled in n^th second =  S_n - S_n-1

S^nth = g/2 [ n² - (n-1)² ]

      = g/2 [ (2n-1) ]  .......................1

this is the formula for finding distance travelled in n^th second....

distance travelled during fifth second is S^5th at n = 5

 S^5th = g/2[ (2*5-1)] = 9g/2                 (g  = 10m/s²)

                               =45m   

approve my ans if u like it

Here's a formula for the distance traversed by the body in nth second:

Snth = u + a/2 (2n-1)

u = intial velocity of the body,

a = acceleration of the body (with proper algebric signs)

you can also apply the same formula for displacement by using vector addition and substraction formula.