A cat moves uniformly with speed 5 ms^-1, so that it is always headed towards a rat is moving on a straight line with constant velocity 3 ms^-1. At initial moment, their velocities are perpendicular to each other and they are separated a distance 16 m apart. Find the time (in second) after which cat catches the rat.

The relative velocity of approach of cat to rat at point P is  $-\frac{dl}{dt}=v_0-v_r\cos \theta =5-3\cos \theta$

$\begin{array}{l}\Rightarrow {\int }_{16}^0 -dl={\int }_0^{t_0} 5dt-3{\int }_0^{t_0} \cos \theta dt\\ \Rightarrow 16=5t_0-3{\int }_0^{t_0} \cos \theta dt ......\left(i\right)\end{array}$

For catching the rat by cat, horizontal distance travelled by both should be same

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}_r{t}_0={\int }_0^{t_0} v_c\cos \theta dt\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3t_0=5{\int }_0^{t_0} \cos \theta dt\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\int }_0^{t_0} \cos \theta dt=\frac{3}{5}{t}_0 .......\left(ii\right)\end{array}$

From Eqs. (i) and (ii), we get

$\begin{array}{l}16=5t_0-3\times \frac{3}{5}{t}_0=5t_0-\frac{9}{5}{t}_0=\frac{16}{5}{t}_0\\ \therefore t_0=5\mathrm{s}\end{array}$