A particle describing a uniform accelerated motion covers 110 m in 2s and 560 m in 8s from the start. Calculate initial velocity and uniform acceleration.

Explanation

Given `" "x_(1)=110m, t_(1)=2s`
`x_(2)=560m, t_(2)=8s`
We know that `x=v_(0)t+1//2at^(2)`
hence `" "110=v_(0)(2)+1//2a(2)^(2)" …………..(1)"`
`560=v_(0)(8)+1//2a(8)^(2)" ………...(2)"`
`"i.e. "560=8v_(0)+32a div 8`
`70=v_(0)+4a" ................(3)"`
`110 =2v_(0)+2a div 2" ...............(4)"`
`55=v_(0)+a`
(3) - (4) gives
`15=3a`
`therefore" "a=5ms^(-2)`
Using a in (4) we get `55=v_(0)+5`
`therefore" "v_(0)=50ms^(-1)`.