If a balloon lying at an altitude of 30 m fr | vedclass

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

Question

(D) Let the position of the observer be at the origin $(0, 0)$. The balloon is at a constant altitude $y = 30 \ m$. Let its horizontal position at tim

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(D) Let the position of the observer be at the origin $(0, 0)$. The balloon is at a constant altitude $y = 30 \ m$. Let its horizontal position at time $t$ be $x(t)$. Given that the balloon moves horizontally at $1 \ m/s$, we have $x(t) = 1 \cdot t = t$ (assuming it starts at $x=0$ at $t=0$).The distance $s$ of the balloon from the observer is given by $s = \sqrt{x^2 + y^2} = \sqrt{t^2 + 30^2}$.To find the rate at which the balloon is moving away from the observer, we differentiate $s$ with respect to $t$:$\frac{ds}{dt} = \frac{1}{2\sqrt{t^2 + 30^2}} \cdot 2t = \frac{t}{\sqrt{t^2 + 30^2}}$.At $t = 40 \ s$:$\frac{ds}{dt} = \frac{40}{\sqrt{40^2 + 30^2}} = \frac{40}{\sqrt{1600 + 900}} = \frac{40}{\sqrt{2500}} = \frac{40}{50} = 0.8 \ m/s$.Thus, the rate is $0.8 \ m/s$.

If a balloon lying at an altitude of $30 \ m$ from an observer at a particular instant is moving horizontally at the rate of $1 \ m/s$ away from him, then the rate at which the balloon is moving away directly from the observer at the $40^{\text{th}}$ second is (in $m/s$)

Similar Questions

The volume of a cube is increasing at the rate of $8 \, cm^{3}/s$. How fast is the surface area increasing when the length of an edge is $12 \, cm$?

The point on the curve $y^2 = 18x$,at which the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate is . . . . . . . $\left(\frac{dx}{dt} \neq 0\right)$

The equation of motion of a particle is given by $s = at^2 + bt + c$. If the displacement after $1 \ s$ is $20 \ m$,the velocity after $2 \ s$ is $30 \ m/s$,and the acceleration is $10 \ m/s^2$,then which of the following is true?

$A$ particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate.

If by dropping a stone in a quiet lake a wave moves in a circle at a speed of $3.5 \, cm/sec$,then the rate of increase of the enclosed circular region when the radius of the circular wave is $10 \, cm$ is ......... $cm^2/sec$. $\left( \pi = \frac{22}{7} \right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module