JEE Advanced 2023 Paper 1, Physics Section 3 Q5: Motion In One Dimension

Similar triangles. Let $x$ be the person's distance from the lamp post and $y$ the distance of the shadow tip from the lamp post. Similar triangles give $\frac{4}{y}=\frac{1.6}{y-x}$, which simplifies to $4(y-x)=1.6y$, or $2.4y=4x$, i.e. $x=0.6y$.

Speed of shadow tip (ground frame). Differentiating: $\frac{dx}{dt}=0.6\frac{dy}{dt}$. With $\frac{dx}{dt}=60{\text{cm s}}^{-1}$: $\frac{dy}{dt}=100{\text{cm s}}^{-1}$.

Relative to the person. $\frac{dy}{dt}-\frac{dx}{dt}=100-60=40{\text{cm s}}^{-1}$.

Answer: $40{\text{cm s}}^{-1}$.