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?

Explanation

Here, `s_(7)^(th)=20m,s_(9)^(th)=24m, s_(15)^(th)=?`
Let initial velocity is u and acceleration is a.
We know that,
`s_(t)^(th)=u+(1)/(2)a(2t-1)`
`therefore" "s_(7)^(th)=u+(1)/(2)a(2xx7-1)=20`
`u+(1)/(2)a(2xx7-1)=20`
`2u+13a=40" ...(i)"`
`therefore" "S_(9)^(th)=u+(1)/(2)a(2xx9-1)=24`
`u+(1)/(2)a(17)=24`
`2u+17a=48" ...(ii)"`
From equation (i) and (ii),
`{:(2u+17a=48," ...(ii)"),(2u+13a=40," ...(i)"),(-" "-" "-,),(_,):}`
`4a=8rArr a=(8)/(4)=2m//s^(2)`
From equation (i)
`2u+13(2)=40`
`2u+26=40rArr 2u=40-26=14`
`u=7m//s`
Henec, `S_(15)^(th)=u+(1)/(2)a(15xx2-1)`
`=7+(1)/(2)xx2(30-1)=7+29=36m`