A street light is at the top of a 12 m pole. | vedclass

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

Question

(A) Let the man be at a distance $x$ from the pole at any time $t$,and let the height of his shadow on the wall be $y$. The total distance from the po

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(A) Let the man be at a distance $x$ from the pole at any time $t$,and let the height of his shadow on the wall be $y$. The total distance from the pole to the wall is $12 \ m$. When the man is $8 \ m$ from the wall,his distance from the pole is $x = 12 - 8 = 4 \ m$. Given $dx/dt = 1/2 \ m/s$.Using similar triangles,the ratio of the height of the shadow to the height of the man relative to the pole is:$\frac{y}{2} = \frac{12}{x}$$y = \frac{24}{x}$Differentiating with respect to $t$:$\frac{dy}{dt} = -\frac{24}{x^2} \cdot \frac{dx}{dt}$At $x = 4$ and $dx/dt = 1/2$:$\frac{dy}{dt} = -\frac{24}{16} \cdot \frac{1}{2} = -\frac{3}{4} \ m/s$.The negative sign indicates that the length of the shadow is decreasing. Thus,the rate at which the shadow is decreasing is $3/4 \ m/s$.

$A$ street light is at the top of a $12 \ m$ pole. $A$ man $2 \ m$ tall walks away from the pole towards a wall $12 \ m$ away from the pole at a speed of $1/2 \ m/s$. The rate at which his shadow on the wall is decreasing when he is $8 \ m$ from the wall is:

Similar Questions

For the curve $y=5x-2x^3$,if $x$ increases at the rate of $2 \text{ units/sec}$,the rate of change in the slope of the curve at $x=3$ is ......../sec

$A$ cube of ice melts without changing its shape at a uniform rate of $4 \, cm^3/min$. The rate of change of the surface area of the cube,in $cm^2/min$,when the volume of the cube is $125 \, cm^3$,is:

$A$ bullet is shot horizontally and its distance $S$ cm at time $t$ second is given by $S=1200t-15t^2$. Then, the distance covered by the bullet when it comes to rest is: (in $\text{ cm}$)

$A$ body moves according to the formula $v = 1 + t^2$,where $v$ is the velocity at time $t$. The acceleration after $3 \text{ s}$ will be .......... $\text{cm/s}^2$. ($v$ is in $\text{cm/s}$)

The distance covered by a particle in $t$ seconds is given by $x = 3 + 8t - 4t^2$. After $1$ second, the velocity will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module