Question

(II)An ocean fishing boat is drifting just above a school of tuna on a foggy day. Without warning, an engine backfire occurs on another boat 1.55 km away (Fig. 12-33). How much time elapses before the backfire is heard (a) by the fish, and (b) by the fishermen?

Short answer

(a) The time elapses before the backfire is heard by the fish is \(0.993\;{\rm{s}}\).

(b) The time elapses before the backfire is heard by the fishermen is \(4.52\;{\rm{s}}\).

Step-by-step solution

Understanding the concept of speed of moving system

The speed of the moving system is determined by the ratio of distance covered by the moving system to the travelling time.

Given data

The distance between both boats is \(d = 1.55\;{\rm{km}}\).

Evaluating the time elapses before the backfire is heard by the fish

The standard value for the speed of sound in sea water and air is \({v_1} = 1560\;{\rm{m/s}}\) and \({v_2} = 343\;{\rm{m/s}}\) respectively.

The time elapses before the backfire is heard by the fish is calculated below:

\({t_1} = \frac{d}{{{v_1}}}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{t_1} = \frac{{1.55\;{\rm{km}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)}}{{1560\;{\rm{m/s}}}}\\{t_1} = 0.993\;{\rm{s}}\end{aligned}\)

Hence, the time elapses before the backfire is heard by the fish is \(0.993\;{\rm{s}}\).

Evaluating the time elapses before the backfire is heard by the fishermen

The time elapses before the backfire is heard by the fishermen is calculated below:

\({t_2} = \frac{d}{{{v_2}}}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{t_2} = \frac{{1.55\;{\rm{km}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)}}{{343\;{\rm{m/s}}}}\\{t_2} = 4.52\;{\rm{s}}\end{aligned}\)

Hence, the time elapses before the backfire is heard by the fishermen is \(4.52\;{\rm{s}}\).