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A person of height 1.6m is walking away from a lamp post of height 4m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60cms^-1, the speed of the tip of the person's shadow on the ground with respect to the person is ___cm s^-1.

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answer is 40.

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$\begin{array}{l} \tan θ = \frac{4}{x_{1} + x_{2}} = \frac{1.6}{x_{2}} \\ \frac{5}{2} (x_{2} - x_{1}) - x_{2} \\ \frac{3}{2} x_{2} = x_{1} \end{array}$
Differentiating it
$\frac{3}{2} \frac{{dx}_{2}}{dt} = \frac{{dx}_{1}}{dt} \frac{{dx}_{2}}{dt} = \frac{2}{3} \times 60 V_{r} = 40$

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