A man of 5 text{ feet} height is walking away | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $H = 15 	ext{ ft}$ be the height of the light and $h = 5 	ext{ ft}$ be the height of the man.Let $x$ be the distance of the man from the lig

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $H = 15 	ext{ ft}$ be the height of the light and $h = 5 	ext{ ft}$ be the height of the man.Let $x$ be the distance of the man from the light source and $s$ be the length of his shadow.By similar triangles, we have $\frac{s}{h} = \frac{x+s}{H}$.Substituting the values, $\frac{s}{5} = \frac{x+s}{15} \implies 3s = x + s \implies 2s = x$.Differentiating with respect to time $t$, we get $2 \frac{ds}{dt} = \frac{dx}{dt}$.Given $\frac{ds}{dt} = \frac{11}{5} 	ext{ ft/sec}$, so $\frac{dx}{dt} = 2 	imes \frac{11}{5} = \frac{22}{5} 	ext{ ft/sec}$.To convert $\frac{dx}{dt}$ to $	ext{miles/hour}$, we have $\frac{22}{5} 	ext{ ft/sec} = \frac{22}{5} 	imes 3600 	ext{ ft/hour} = \frac{22 	imes 3600}{5 	imes 5280} 	ext{ miles/hour}$.Calculating this, $\frac{79200}{26400} = 3 	ext{ miles/hour}$.Thus, $K = 3$.

$A$ man of $5 \text{ feet}$ height is walking away from a light fixed at a height of $15 \text{ feet}$ at the rate of $K \text{ miles/hour}$. If the rate of increase of his shadow is $\frac{11}{5} \text{ feet/sec}$, then $K=$ (Take $1 \text{ mile} = 5280 \text{ feet}$)

Similar Questions

$A$ point moves along the arc of the parabola $y = 2x^2$. Its abscissa increases uniformly at the rate of $2 \text{ units/sec}$. At the instant the point is passing through $(1, 2)$,its distance from the origin is increasing at the rate of

If the volume of a spherical ball is increasing at the rate of $4 \pi \ cm^3/sec$,then the rate of change of its surface area when the volume is $288 \pi \ cm^3$ is:

The volume of a ball is increasing at the rate of $4 \pi \text{ cc/sec}$. The rate of increase of the radius,when the volume is $288 \pi \text{ cc}$,is

$A$ firm is manufacturing $2000$ items. It is estimated that the rate of change of production $P$ with respect to the additional number of workers $x$ is given by $\frac{dP}{dx} = 100 - 12\sqrt{x}$. If the firm employs $25$ more workers,then the new level of production of items is:

Displacement $s$ of a particle at time $t$ is expressed as $s=2 t^3-9 t$. Find the acceleration at the time when the velocity vanishes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module