Question

The current velocity (V) of a river is directly proportional to the distance (y) from its nearest bank. v becomes
maximum in the middle and is equal to Vo. A boat is moving in the river with a constant velocity u relative to the
water and perpendicular to the current. Find the drift of the boat. Width of the river is d.
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Answer 1

Given that,

Current velocity of river = v

Distance = y

Width = d

Velocity of boat = u

A boat is moving in the river with a constant velocity

$v_{y}=\dfrac{dy}{dt}=u$.....(I)

We need to calculate the velocity of river

Using formula of distance

$v_{x}\propto y$

$v_{x}=ky$....(II)

at $y=\dfrac{d}{2}$  and  $v_{x}=v_{0}$

Put the value in equation (II)

So, The velocity is

$v_{0}=k\times\dfrac{d}{2}$

$k=\dfrac{2v_{0}}{d}$

Put the value of k in equation (II)

$v_{x}=\dfrac{2v_{0}}{d}\times y$

A boat is moving in the river with a constant velocity u relative to the water and perpendicular to the current.

We need to calculate the time

Using formula of velocity

$v_{y}=\dfrac{d}{t}$

$t=\dfrac{d}{u}$

We need to calculate the acceleration in x direction

Using equation (I)

$v_{x}=\dfrac{2v_{0}}{d}y$

On differentiating

$a_{x}=\dfrac{dv_{x}}{dt}=\dfrac{2v_{0}}{d}\dfrac{dy}{dt}$

Put the value of $\dfrac{dy}{dt}$

$a_{x}=\dfrac{2v_{0}u}{d}$

We know that,

Drift :

Drift is the distance between two points.

We need to calculate the drift of the boat

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$x=0+\dfrac{1}{2}\times\dfrac{2v_{0}u}{d}\times(\dfrac{d}{u})^2$

$x=\dfrac{v_{0}d}{u}$

Hence, The drift of the boat is $\dfrac{v_{0}d}{u}$