A man 'A' moves in north direction with speed 10m/s and another man B moves in `E-30^(@)-N` with `10 m//s`. Find the relative velocity of `B w.r.t. A.

Explanation

The speed v, tangential acceleration `a_(tau)` and the normal acceleration `a_(n)` are expressed bu the following equations.
`v=(ds)/(dt)`
Substituting the given expression for s, we have `v=2t^(2)`
`a_(tau)=(d^(2)s)/(dt^(2))` …(i)
Substituting the given expression for s, we have `a_(tau)=4t`
`a_(n)=v^(2)/r` ...(ii)
Substituting v from eq. (i) and `r=8 m`, we have `a_(n)=1/2 t^(4)` ...(iii)
The instant when the tangential and the normal accelerations have equal magnitude, can be obtained by equating their expressions given in eq. (ii) and (iii).
`a_(n)=a_(T) rarr " "t=2s`
Substituting the above value of t in eq. (i), we obtain `v=8 m//s`