Question

A dolphin echolocates its prey in $1.25\text{s}$. Assuming the prey is still, how long does it take for the dolphin to reach it, if it starts travelling at an average speed of $15\text{kmph}$? (Take the speed of sound in water to be $1400\text{m}{\text{s}}^{-1}$)

• A. $2.0\text{min}$
• B. $4.5\text{min}$
• C. $5.0\text{min}$
• D. $3.5\text{min}$

Answer 1

The correct option is D $3.5\text{min}$

Given:
The speed of sound in water, $\text{s}=1400\text{m}{\text{s}}^{-1}$
Time taken by the dolphin to locate its prey using sound, $\text{t}=1.25\text{s}$
The sound from the dolphin goes to the prey, hits it, and reflects back to the dolphin in $\text{t}=1.25\text{s}$
Also the sound travels twice the distance $\left(2d\right)$ between the dolphin and its prey, in the time period $\text{t}$.
$⟹$ The distance to the prey, $\text{d}=\frac{\text{st}}{2}$
$⟹\text{d}=\frac{1400\times 1.25}{2}=875\text{m}$
Also given:
The average speed of the dolphin, ${\text{v}}_{\text{d}}=15\text{kmph}=15\times \frac{5}{18}\text{m}{\text{s}}^{-1}$
${\text{v}}_{\text{d}}=\frac{25}{6}\text{m}{\text{s}}^{-1}$
$\therefore$ the time taken by the dolphin to reach its prey $=\frac{\text{d}}{{\text{v}}_{\text{d}}}=\frac{875}{\left(\frac{25}{6}\right)}=\frac{875\times 6}{25}=210\text{s}$

$210\text{s}=\frac{210}{60}\text{min}=\frac{7}{2}\text{min}=3.5\text{min}$