Physics · Motion in one Dimension

JEE Advanced 2023 — Paper 1 — Question 28

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 $60 \mathrm{~cm} \mathrm{~s}^{-1}$ , The speed of the tip of the person's shadow on the ground with respect to the person is $\qquad$ $\mathrm{cm} \mathrm{s}{ }^{-1}$

Answer: 40

Numerical answer — enter this value.

Step-by-step solution

$\frac{4}{x_{2}}=\frac{1.6}{\left(x_{2}-x_{1}\right)}$ $\Rightarrow 3 x_{2}=5 x_{1}$ $\Rightarrow 3 \frac{\mathrm{dx}_{2}}{\mathrm{dt}}=5 \frac{\mathrm{dx}_{1}}{\mathrm{dt}}$ $\Rightarrow \frac{\mathrm{dx}_{2}}{\mathrm{dt}}=\frac{5}{3} \times 60=100 \mathrm{~cm} / \mathrm{s}$ $\Rightarrow \mathrm{V}_{\text {rel }}=40 \mathrm{~cm} / \mathrm{sec}$

Answer key and solution verified before publishing.

Practise Motion in one Dimension

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion