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2.Motion in Straight Line
easy
A person of height $1.6 \mathrm{~m}$ is walking away from a lamp post of height $4 \mathrm{~m}$ 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. . . . $\mathrm{cm} \mathrm{s}^{-1}$.
A
$20$
B
$30$
C
$40$
D
$50$
(IIT-2023)
Solution
$\frac{4}{y}=\frac{1.6}{y-x}$
$4 y-4 x=1.6 y$
$2.4 y=4 x$
$x=0.6 y$
$\frac{d x}{d t}=0.6 \times \frac{d y}{d t}$
$60=0.6 \times \frac{d y}{d t}$
$\therefore \frac{d y}{d t}=100 \mathrm{~cm} / \mathrm{s}$
Speed of tip of person's
Shadow $w.r.t$ person $=100-60=40 \mathrm{~cm} / \mathrm{s}$
Standard 11
Physics
