Animated Solution for Physics - Kinematics: A person of height $1.6\text{ m}$ is walking away from a lamp post of height $4\text{ m}$ 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 $60{\text{ cm s}}^{-1}$, the speed of the tip of the person's shadow on the ground with respect to the person is _______ ${\text{cm s}}^{-1}$.

Imagine walking down a quiet street at night. As you walk away from a streetlamp, you might have noticed your shadow stretching out in front of you, moving faster than you are. Have you ever wondered exactly how fast it moves? This beautiful problem from JEE Advanced 2023 takes that everyday experience and turns it into a fascinating exercise in geometry and kinematics.

The Geometry of Shadows

Let's translate this physical reality into a mathematical model. We have a lamp post standing tall at $4\text{ m}$, and a person of height $1.6\text{ m}$ walking away from it. The light ray from the lamp grazes the top of the person's head and hits the ground, creating the tip of the shadow.

Let the distance of the person from the lamp post be $x$, and the distance of the shadow's tip from the lamp post be $y$. The length of the shadow itself is simply the difference between these two, which is $y-x$.

Here is where the magic happens. The light ray forms two right-angled triangles that share the same angle at the tip of the shadow. The larger triangle is formed by the lamp post and the total distance $y$. The smaller triangle is formed by the person and the shadow length $y-x$.

Because these triangles are similar, the ratio of their heights must equal the ratio of their bases. We can write this elegantly as:

Let's simplify this relation. Cross-multiplying gives us:

Expanding the terms, we get $4y-4x=1.6y$. Rearranging this to isolate our variables yields $2.4y=4x$, which simplifies beautifully to:

This simple linear equation is the geometric heart of our problem. It tells us that the person's position is always exactly $60\%$ of the shadow tip's position.

The Magic of Calculus

Now, we need to bring motion into the picture. How do we transition from static positions to dynamic speeds? We differentiate with respect to time!

Taking the time derivative of our position equation $x=0.6y$, we get:

Here, $\frac{dx}{dt}$ is the rate at which the person's position is changing—that's the person's speed, $v_p$. Similarly, $\frac{dy}{dt}$ is the speed of the shadow's tip with respect to the ground, $v_s$.

We are given that the person is walking at $v_p=60\text{ cm/s}$. Substituting this into our derivative equation:

Solving for $v_s$, we find:

So, the tip of the shadow is zooming across the ground at $100\text{ cm/s}$!

Relative Velocity

If you stop here, you might fall into a classic trap. The question doesn't ask for the shadow's speed with respect to the ground. It specifically asks for the speed of the shadow's tip with respect to the person.

Since both the person and the shadow are moving in the same direction, we need to find their relative velocity. We do this by subtracting the person's speed from the shadow's speed:

And there we have it! The tip of the shadow is moving away from the person at a relative speed of $40\text{ cm/s}$. This problem is a brilliant reminder of how simple geometric principles, when combined with basic calculus, can perfectly describe the dynamic world around us.