Question

A person of height 1.6 m is walking away from a lamp post of height 4 m 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 60 cm s^−1, the speed of the tip of the person’s shadow on the ground with respect to the person is _______ cm s^−1.

Answer 1

Given that \(\frac{dx1}{dt}\) = speed of person = 60 cm/s 

Shadow Speed Problem 

Also, \( \frac{dx_2}{dt} \) is the speed of the tip of the person's shadow.

Applying a similar triangle rule in \( \triangle ABE \) and \( \triangle DCE \):

\[ \frac{4}{x_2} = \frac{1.6}{x_2 - x_1} \]

This simplifies to: \[ 2x_2 - 4x_1 = 1.6x_2 \]

Rearranging the equation: \[ 2.4x_2 = 4x_1 \] 

Now, differentiating both sides with respect to \( t \):

\[ 2.4 \frac{dx_2}{dt} = 4 \frac{dx_1}{dt} \]

Given that \( \frac{dx_1}{dt} = 100 \, \text{cm/s} \), we can substitute this into the equation:

\[ 2.4 \frac{dx_2}{dt} = 4 \times 100 \, \text{cm/s} \]

Thus, the speed of the tip of the shadow is:

\[ V_{\text{SP}} = 40 \, \text{cm/s} \]