Ship Rescue Operation
Given Data:
Velocity of ship $v_s = 36 \text{ km/h} = 10 \text{ m/s}$.
Velocity of rescue boat $v_b = 72 \text{ km/h} = 20 \text{ m/s}$.
Length of ship $l = 150 \text{ m}$.
At encounter (front of ship): Sinking boat distance $x_0 = 3.0 \text{ km} = 3000 \text{ m}$.
Rescue time $t_0 = 1.0 \text{ min} = 60 \text{ s}$.
Phase 1: Boat travels from Mid-Ship to Leading Edge
The boat starts at the middle of the ship. Relative to the ship, it must cover distance $l/2 = 75$ m.
Relative velocity $v_{rel1} = v_b – v_s = 20 – 10 = 10$ m/s.
Time taken: $t_1 = \frac{75}{10} = 7.5$ s.
Phase 2: Boat travels to Sinking Boat
At the instant the boat passes the leading edge (time $t_1$), the sinking boat is $x_0 = 3000$ m away.
Time to cover this distance: $t_2 = \frac{x_0}{v_b} = \frac{3000}{20} = 150$ s.
Phase 3: Rescue Operation
Time spent at the site: $t_3 = t_0 = 60$ s.
Phase 4: Return Journey
Total time elapsed before return starts: $T_{out} = t_1 + t_2 + t_3 = 7.5 + 150 + 60 = 217.5$ s.
Now, we calculate the separation between the rescue boat and the ship’s mid-point at this instant.
Let the starting position of the ship’s mid-point be $x=0$ at $t=0$.
Position of Ship’s Mid-point at $T_{out}$: $X_{ship} = v_s T_{out} = 10 \times 217.5 = 2175$ m.
Position of Rescue Boat (Sinking boat location):
We find the total distance the boat traveled from the origin. The boat reached the front ($t=7.5$) then traveled 3000m. But the reference frame is tricky. Let’s use coordinates.
At $t=7.5$, Boat is at leading edge. Leading edge position = $X_{mid}(7.5) + 75 = 10(7.5) + 75 = 150$ m.
From there, it traveled 3000m to the sinking boat. So Sinking Boat Position $X_{sink} = 150 + 3000 = 3150$ m.
Separation at $T_{out}$: $\Delta x = 3150 – 2175 = 975$ m.
Relative velocity for return (Approaching): $v_{rel2} = v_b + v_s = 20 + 10 = 30$ m/s.
Time to return: $t_4 = \frac{975}{30} = 32.5$ s.
Total Time
$T_{total} = T_{out} + t_4 = 217.5 + 32.5 = 250$ s.
Answer: 250 s