Physics, asked by apple283jjj, 11 months ago

a person can swim in still water with velocity 10m/minute if the river flows with the velocity 4m/minute and width of the river is 500m he can cross the river in the shortest time of

Answers

Answered by chetlasrijith

Hey mate here is your answer please mark as brainlist

he should swim due north!!!

his velocity in moving water is equal to his own velocity plus the velocity of the river.

now to cross the river in smallest possible time he should maximise the velocity in the north direction.

suppose he move at an angle of @ from the bank then the component due north is 10sin@m/min. because river velocity is zero in the north direction.

now this velocity due noth will be maximum when @ tends to 90 degrees.

hence he should move due north that is 90 degrees to the bank.

Answered by Laxmipriyas007

Answer:

He crosses the river in the shortest time of 50 minutes.

Explanation:

Given: A person can swim in still water at 10 m/minute the river flows at 4 m/minute.

To find: If he crosses the river in the shortest time ------------?

Let the velocity of man be V2 and the velocity of the river is V1.

We know that the resultant velocity is $\sqrt{V2^2 + V1^2}$

= $\sqrt{10^2 + 4^2}$

= $\sqrt{100 + 16}$

= $\sqrt{116}$ m/minute

Let 'X' be the angle from the vertical along the direction of the flow

Cos (x) = $\frac{10}{\sqrt{116}}$

Let 'L' be the width of the river and H be the distance the swimmer will travel

$\frac{L}{H}$ = $\frac{10}{\sqrt{116}}$

$\frac{H}{L}= \frac{1}{cos(x)}=\frac{\sqrt{116}}{{10}}$

${H}=500\times \frac{\sqrt{116}}{{10}}$

${H}=50\times \sqrt{116}$ (since, the width of the river is 500m=L)

Time the person takes to swim =(50/✓116)/✓116

Time, therefore, is 50 minutes.

The man will reach 50*4m = 200m away from the opposite side of the point from where he starts from.