Question

A boat man finds that he can save 8 s in crossing a river by quickest path than by the shortest path. If the velocity of the boat and the river flow are 13 m/s and 12 m/s respectively, then the width of the river is:

• A. 65 m
• B. 56 m
• C. 75 m
• D. 57 m

Hint:

For problems involving motion in rivers, remember to consider the effective velocity perpendicular to the current when calculating times.

Answer 1

The Correct Option is A

Let the velocity of the boat be \( v_b = 13 \, {m/s} \) and the velocity of the river be \( v_r = 12 \, {m/s} \). The time saved by following the quickest path is given as \( \Delta t = 8 \, {s} \). 
Step 1: Without the river current, the time to cross the river would be: \[ t_1 = \frac{d}{v_b} \] Step 2: With the river current, the effective velocity of the boat across the river is \( \sqrt{v_b^2 - v_r^2} \). The time to cross the river is: \[ t_2 = \frac{d}{\sqrt{v_b^2 - v_r^2}} \] Step 3: The time saved is the difference between these times: \[ \Delta t = t_1 - t_2 = 8 \] Substituting the given values \( v_b = 13 \, {m/s} \) and \( v_r = 12 \, {m/s} \): \[ \frac{d}{13} - \frac{d}{\sqrt{13^2 - 12^2}} = 8 \] \[ \frac{d}{13} - \frac{d}{\sqrt{25}} = 8 \] \[ \frac{d}{13} - \frac{d}{5} = 8 \] Multiplying both sides by 65: \[ 5d - 13d = 8 \times 65 \] \[ -8d = 520 \] \[ d = 65 \, {m} \] Thus, the width of the river is 65 meters, i.e., option (A).