If in the previous problem, particle `A` moves with constant acceleration `4 m//s^(2)` with initial speed `5 m//s` and `B` moves with uniform speed `12 m//s`, when and where the particle meet?

Explanation

For `A:d =5t+1/2xx4t^(2)`
`=5t+2t^(2)` (accelerated motion) …(i)
For `B: 100-d=12t` (uniform motion) …(ii)
Adding (i) and (ii), we get
`100=2t^(2)+17t`
`2t^(2)+17t-100=0`
`t=(-17+-sqrt((17)^(2)+4(2)(-100)))/(2xx2)`
`=(-17+-sqrt(289+800))/4`
`=(-17+-33)/4=4 s, d=52m`