Solution 18: River Crossing Kinematics
Given Parameters:
- Time to cross minimizing time: $t_{min} = 60$ min = 1 hr.
- Time to cross minimizing drift (zero drift): $t_{drift=0} = 180$ min = 3 hr.
- Let velocity of swimmer relative to water be $v_b$ and velocity of river be $v_r$. Width of river is $d$.
Fig 1: Case 1 (Left) is Minimum Time. Case 2 (Right) is Zero Drift.
Analysis
Case 1: Minimum Time
To cross in minimum time, the swimmer swims perpendicular to the river flow.
$$ t_1 = \frac{d}{v_b} = 1 \text{ hr} \implies d = v_b $$
Case 2: Minimum Drift (Zero Drift)
To have zero drift, the swimmer must swim upstream at an angle such that the horizontal component cancels the river flow ($v_x = v_r$). The resultant velocity across the river is $\sqrt{v_b^2 – v_r^2}$.
$$ t_2 = \frac{d}{\sqrt{v_b^2 – v_r^2}} = 3 \text{ hr} $$
Solving the system:
Substitute $d = v_b$ into the second equation:
$$ 3 = \frac{v_b}{\sqrt{v_b^2 – v_r^2}} $$
Squaring both sides:
$$ 9 (v_b^2 – v_r^2) = v_b^2 $$
$$ 9v_b^2 – 9v_r^2 = v_b^2 \implies 8v_b^2 = 9v_r^2 $$
$$ v_b = \frac{3}{2\sqrt{2}} v_r \approx 1.06 v_r $$
Evaluating Options
- (a) Since $v_b \approx 1.06 v_r$, $v_b > v_r$. True.
- (b) Contradicts (a). False.
- (c) Given $d = 3\sqrt{2}$ km and $v_r = 4$ km/h.
From $t_1 = 1$, $v_b = d/1 = 3\sqrt{2}$ km/h.
Check relation: $v_b = \frac{3}{2\sqrt{2}} v_r = \frac{3}{2\sqrt{2}}(4) = 3\sqrt{2}$. Matches. True. - (d) Crossing a $3\sqrt{2}$ km river in $60\sqrt{2}$ min ($t = \sqrt{2}$ hr).
Velocity across river $v_y = \frac{d}{t} = \frac{3\sqrt{2}}{\sqrt{2}} = 3$ km/h.
Using $v_b = 3\sqrt{2}$, find horizontal component $v_x$:
$v_b^2 = v_x^2 + v_y^2 \implies (3\sqrt{2})^2 = v_x^2 + 3^2 \implies 18 = v_x^2 + 9 \implies v_x = 3$ km/h.
Swimmer swims upstream with $v_x = 3$. River flows downstream with $v_r = 4$.
Net drift velocity = $v_r – v_x = 4 – 3 = 1$ km/h.
Drift = Velocity $\times$ time = $1 \text{ km/h} \times \sqrt{2} \text{ h} = \sqrt{2}$ km. True.
Correct Options: (a), (c), and (d)