A force of `2N` acts for `5 seconds` on a particle of mass `0.5 kg` initially at rest. Calculate the distance moved by the particle in (i) these five seconds and (ii) next `5 seconds`

Explanation

Here, `F=2N, t=5s, m= 0.5kg, u-0`
`a=F/m= 2/(0.5)= 4m//s^(2)`
(i) `s = ut+1/2at^(2)= 0+1/2xx4(5)^(2)= 50` metre.
(ii) velocity acquired at the end of 5 seconds is
`v= u+at`
`v=0+4xx5= 20m//s`.
This velocity is uniformed, i.e., `a=0` as no force is being applied.
`:.` Distance mov ed in next 5 seconds
`s' = vxxt= 20xx5 = 100m`