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

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(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 $m/s$)

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