KINEMATICS BYU 26 - Physics.lt

Solution

Given:
Length of ship: \[ l = 150\,\text{m} \] Speed of ship: \[ v_s = 36\,\text{km h}^{-1} = 10\,\text{m s}^{-1} \] Speed of rescue boat: \[ v_b = 72\,\text{km h}^{-1} = 20\,\text{m s}^{-1} \] Time spent at the sinking boat: \[ t_0 = 1\,\text{min} = 60\,\text{s} \]

Key observation: The rescue boat is lowered from the middle of the \(150\,\text{m}\) long ship. Hence initially it is \(75\,\text{m}\) behind the leading edge of the ship.

1. Time to overtake the leading edge of the ship

Both the ship and the rescue boat initially move in the same direction. Therefore, the relative speed of the rescue boat with respect to the ship is

\[ v_{\text{rel}} = v_b-v_s =20-10 =10\,\text{m s}^{-1} \]

The rescue boat has to gain a distance of \(75\,\text{m}\).

\[ t_a =\frac{75}{20-10} =7.5\,\text{s} \]

2. Time to reach the sinking boat

At the instant the rescue boat crosses the leading edge of the ship, the sinking boat is given to be \(3.0\,\text{km}\) ahead.

\[ t_b =\frac{3000}{20} =150\,\text{s} \]

Hence the total outward journey takes

\[ t_{\text{out}} =t_a+t_b =7.5+150 =157.5\,\text{s} \]

3. Separation from the ship when the return journey begins

Consider the instant at which the rescue boat overtakes the leading edge of the ship.

At this instant, the midpoint of the ship is \(75\,\text{m}\) behind its leading edge, while the sinking boat is \(3000\,\text{m}\) ahead of the leading edge.

Therefore, the separation between the ship’s midpoint and the sinking boat is

\[ 3000+75=3075\,\text{m} \]

From this instant until the rescue boat starts its return journey, the elapsed time is

\[ \frac{3000}{20}+60 =150+60 =210\,\text{s} \]

During these \(210\,\text{s}\), the midpoint of the ship moves forward by

\[ 10(210)=2100\,\text{m} \]

Thus, when the rescue boat starts returning, its separation from the midpoint of the ship is

\[ d=3075-2100 =975\,\text{m} \]

4. Return journey

Now the rescue boat moves towards the ship with speed \(20\,\text{m s}^{-1}\), while the ship moves towards the approaching boat with speed \(10\,\text{m s}^{-1}\).

Therefore, their closing speed is

\[ v_{\text{closing}} =20+10 =30\,\text{m s}^{-1} \]

Hence the return time is

\[ t_{\text{return}} =\frac{975}{30} =32.5\,\text{s} \]

5. Total time of the rescue operation

\[ \begin{aligned} T &=t_{\text{out}}+t_0+t_{\text{return}}\\[4pt] &=157.5+60+32.5\\[4pt] &=250\,\text{s} \end{aligned} \]

Therefore, \[ \boxed{T=250\,\text{s}} \] or \[ \boxed{T=4\,\text{min}\;10\,\text{s}} \]