A particle describing a uniform accelerated motion, covers 118 m in the 5th second and 138 m in 10th second. Calculate the initial velocity and acceleration of the particle. What distance will be covered by the particle at the end of 20 s?

Explanation

We know that `s_(n^("th"))=v_(0)+(a)/(2)(2n-1)`
`"given "S_(n^("th"))=118m,s_(10th)=138m`
`"i.e. "118=v_(0)+(a)/(2)(2xx5-1)=v_(0)+(9)/(2)a" ……………….(1)"`
`"and "138=v_(0)+(a)/(2)(2xx10-1)=v_(0)+(19a)/(2)`
(2)-(1) gives
`20=(a)/(2)(19-9)=5a`
`therefore" "a=4ms^(-2)`
Using a in (1) we get
`118 =v_(0)+(9)/(2)xx4`
`v_(0)=118-18=100ms^(-1)`
Distance covered at the end of 20 s by applying `x=v_(0)t+1//2at^(2)`
`"i.e. "x=100xx20+(1)/(2)xx4xx20xx20=2000+800=2800m`
or distance covered = 2.8km