A particle travels a distance of 20 m in the `7^(th)` seconds and 24 m in `9^(th)` second. How much distance shall it travel in the `15^(th)` second?

Options

A. 10 m
B. 16 m
C. 24 m
D. 36 m

Answer

D

Explanation

Here, `D_7 = 20 m , D_(9) = 24 m , D_(15) = ? `
Let u and a be the initial velocity and uniform acceleration the body
`D_(n) =u+a/2 (2n-1)`
`:. D_(7) = u + a/2(2xx7-1) " or " 20 = u +(13a)/2 " "....(i)`
and `D_(9) =u +a/2 (2xx9 -1) " or " 24 =u +(17)/(2)a " "...(ii)`
Subtracting equation (i) from equation (ii) , we get
`4=2a " or " 2 ms^(-2)`
Putting this value in equation (i),we ge `20 = u +(13)/2 xx2 = u +13 " or " u = 7 ms ^(-1)`
`:. D_(15) = u +a/2 (2xx15-1) =7+2/2xx29 = 36 m`