Question

A river is flowing with velocity $5km/hr$ as shown in the figure. A boat starts from $A$ and reaches the other bank by covering shortest possible distance. Velocity of boat in still water is $3km/hr$ . The distance boat covers is :

• A. $500m$
• B. $400\sqrt{2}$
• C. $300\sqrt{2}$
• D. $600m$

Answer 1

The correct option is A $500m$

Let the velocity of boat making an angle $\theta$ with the line AB.

The time to cross the breadth AB is given as,

$t=\frac{AB}{v_{AB}}$

$t=\frac{0.3}{3\cos \theta }$

$t=0.1\sec \theta$

The actual velocity of the boat is given as,

$v=\sqrt{3^2+5^2+2\times 3\times 5\cos \left({90}^{\circ }+\theta \right)}$

$v=\sqrt{34-30\sin \theta }km/h$

The distance travelled to cross the river is given as,

$s=vt$

$s=\sqrt{34-30\sin \theta }\times 0.1\sec \theta$

$100s^2=\left(34-30\sin \theta \right)\times {\sec }^2\theta$

Differentiate the above expression with respect to $\theta$, we get

$200s\frac{ds}{d\theta }=\left(34-30\sin \theta \right)\times 2{\sec }^2\theta tan\theta -30\cos \theta \times {\sec }^2\theta$

For minimization $\frac{ds}{d\theta }$ should be zero and it can be written as,

$0=\left(34-30\sin \theta \right)\times 2{\sec }^2\theta tan\theta -30\cos \theta \times {\sec }^2\theta$

$15{\sin }^2\theta -34\sin \theta +15=0$

$\sin \theta =\frac{34-\sqrt{{\left(34\right)}^2-4\times 15\times 15}}{30}$

$\sin \theta =\frac{3}{5}$

$\sec \theta =\frac{5}{4}$

The distance is given as,

$s=\sqrt{34-30\times \frac{3}{5}}\times 0.1\times \frac{5}{4}$

$s=0.5km$ or $500m$

The possible shortest distance is $500m$.