# Question #aea85

Since the boat crosses the river of width 1km along the shotest possible path in 15min or 0.25hr, its resultant velocity of crossing the river will be ${V}_{\text{cross"=1/0.25=4" km/hr}}$.This velocity will be perpendicular to the direction of the velocity of river ${V}_{\text{river}}$. Given the velocity of boat in still water ${V}_{\text{boat"=5" km/hr}}$
${V}_{\text{boat"^2=V_"river"^2+V_"cross}}^{2}$
$\implies {5}^{2} = {V}_{\text{river}}^{2} + {4}^{2}$
${V}_{\text{river"=sqrt(5^2-4^2)=3" km/hr}}$