A particle travels m a straight line, such that for a short time `2 s le tle 6 s,` its motion is described by `v= (4/a) m//s,` where a is in `m//s^2.` If `v= 6 m//s.` when `t= 2 s,` determine the particle's acceleration when `t= 3 s.`

Answer

B, C

Explanation

`a=4/v or (dv)/(dt)=4/v`
`:. int_6^v vdv=int_2^t4 dt`
`:. v^2/2-18=4t-8`
`:. v=sqrt(8t+20)`
`a=(dv)/(dt)=4/(sqrt(8t+20))`
`At t=3s`
`a=0.603 m//s^2`