Question

A man wants to reach point B on the opposite bank of a river flowing at a speed 4 m/s starting from A as shown in figure. What minimum speed relative to water should the man have so that he can reach point B directly by swimming?

• A. $\frac{4}{5}$ m/s
• B. $\frac{12}{5}$ m/s
• C. $\frac{16}{5}$ m/s
• D. $\frac{23}{5}$ m/s

Answer 1

Answer: (C)

$\frac{16}{5}$ m/s

Answer 2

To solve this problem, we need to understand the concept of relative velocity. The velocity of the man relative to the ground ($\vec{v}_g$) is the vector sum of his velocity relative to the water ($\vec{v}_m$) and the velocity of the river ($\vec{v}_r$).

So, $\vec{v}_g = \vec{v}_m + \vec{v}_r$.

Let's set up a coordinate system:

• Let the starting point A be the origin (0, 0).
• The river flows in the positive x-direction. So, the velocity of the river is $\vec{v}_r = 4 \hat{i}$ m/s.
• The width of the river is 40 m, so the opposite bank is at y = 40 m.
• Point B is 30 m upstream from the point directly opposite A. This means point B is at coordinates (-30, 40) m.

The man wants to reach point B directly from A. This implies that his resultant velocity relative to the ground ($\vec{v}_g$) must be directed along the line segment AB.

The displacement vector from A to B is $\vec{AB} = (-30 - 0) \hat{i} + (40 - 0) \hat{j} = -30 \hat{i} + 40 \hat{j}$.

The direction of $\vec{v}_g$ must be parallel to $\vec{AB}$.

So, $\vec{v}_g = k \vec{AB}$ for some scalar $k > 0$.

Let $\vec{v}_g = v_{gx} \hat{i} + v_{gy} \hat{j}$.

The ratio of components of $\vec{v}_g$ must be equal to the ratio of components of $\vec{AB}$:

$\frac{v_{gy}}{v_{gx}} = \frac{40}{-30} = -\frac{4}{3}$.

So, $3 v_{gy} = -4 v_{gx}$, or $4 v_{gx} + 3 v_{gy} = 0$.

Let the velocity of the man relative to water be $\vec{v}_m = v_{mx} \hat{i} + v_{my} \hat{j}$.

We want to find the minimum magnitude of $\vec{v}_m$, i.e., minimize $v_m = \sqrt{v_{mx}^2 + v_{my}^2}$.

Using the relative velocity equation: $\vec{v}_g = \vec{v}_m + \vec{v}_r$.

Substituting the components:

$(v_{gx} \hat{i} + v_{gy} \hat{j}) = (v_{mx} \hat{i} + v_{my} \hat{j}) + (4 \hat{i})$

Equating the components:

$v_{gx} = v_{mx} + 4$ (Equation 1)

$v_{gy} = v_{my}$ (Equation 2)

Substitute $v_{gx}$ and $v_{gy}$ into the direction condition $4 v_{gx} + 3 v_{gy} = 0$:

$4 (v_{mx} + 4) + 3 v_{my} = 0$

$4 v_{mx} + 16 + 3 v_{my} = 0$

$4 v_{mx} + 3 v_{my} = -16$ (Equation 3)

We need to minimize $v_m = \sqrt{v_{mx}^2 + v_{my}^2}$. This is equivalent to minimizing $v_m^2 = v_{mx}^2 + v_{my}^2$.

From Equation 3, $v_{my} = -\frac{4}{3} v_{mx} - \frac{16}{3}$.

Substitute this into the expression for $v_m^2$:

$v_m^2 = v_{mx}^2 + \left(-\frac{4}{3} v_{mx} - \frac{16}{3}\right)^2$

$v_m^2 = v_{mx}^2 + \frac{1}{9} (4 v_{mx} + 16)^2$

$v_m^2 = v_{mx}^2 + \frac{1}{9} (16 v_{mx}^2 + 128 v_{mx} + 256)$

$v_m^2 = \frac{9 v_{mx}^2 + 16 v_{mx}^2 + 128 v_{mx} + 256}{9}$

$v_m^2 = \frac{25 v_{mx}^2 + 128 v_{mx} + 256}{9}$

To find the minimum value of $v_m^2$, we can find the vertex of this quadratic expression in $v_{mx}$. For a quadratic $ax^2 + bx + c$, the minimum (or maximum) occurs at $x = -b/(2a)$.

Here, $a = 25/9$, $b = 128/9$.

So, $v_{mx}$ for minimum $v_m$ is:

$v_{mx} = -\frac{128/9}{2 \times (25/9)} = -\frac{128}{50} = -\frac{64}{25}$ m/s.

Now, substitute this value of $v_{mx}$ back into Equation 3 to find $v_{my}$:

$4 \left(-\frac{64}{25}\right) + 3 v_{my} = -16$

$-\frac{256}{25} + 3 v_{my} = -16$

$3 v_{my} = -16 + \frac{256}{25}$

$3 v_{my} = \frac{-400 + 256}{25}$

$3 v_{my} = -\frac{144}{25}$

$v_{my} = -\frac{144}{3 \times 25} = -\frac{48}{25}$ m/s.

Finally, calculate the minimum speed relative to water, $v_m$:

$v_m = \sqrt{v_{mx}^2 + v_{my}^2}$

$v_m = \sqrt{\left(-\frac{64}{25}\right)^2 + \left(-\frac{48}{25}\right)^2}$

$v_m = \sqrt{\frac{4096}{625} + \frac{2304}{625}}$

$v_m = \sqrt{\frac{4096 + 2304}{625}}$

$v_m = \sqrt{\frac{6400}{625}}$

$v_m = \frac{\sqrt{6400}}{\sqrt{625}} = \frac{80}{25}$

$v_m = \frac{16 \times 5}{5 \times 5} = \frac{16}{5}$ m/s.