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A drone travels vertically from its launch pad.
It's height above ground, \(h\) metres, at time \(t\) minutes is modelled by
\(h(t)=-0.2 t^3+3 t^2+5 t\) for \(0 \leq t \leq 12\)
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a. \(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)
b. \(\dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)
a. \(h=-0.2 t^3+3 t^2+5 t\)
\(\text{Velocity of the drone}=\dfrac{d h}{d t}.\)
\(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)
b. \(\text{Drone is descending when} \ \ \dfrac{dh}{dt}<0:\)
| \(-0.6 t^2+6 t+5\) | \(<0\) | |
| \(0.6 t^2-6 t-5\) | \(>0\) | |
| \(6 t^2-60 t-50\) | \(>0\) |
\(\text{Solve}\ \ 6 t^2-60 t-50=0:\)
\(t=\dfrac{60 \pm \sqrt{(-60)^2+4 \times 6 \times 50}}{2 \times 6}=\dfrac{60 \pm \sqrt{4800}}{12}=\dfrac{15 \pm 10 \sqrt{3}}{3}\)
\(\text{Since parabola is concave up:}\)
\(6 t^2-60 t-50>0\ \ \text{when}\ \ t>\dfrac{15+10 \sqrt{3}}{3} \quad\left( t=\dfrac{15-10 \sqrt{3}}{3}<0\right)\)
\(\therefore \text{Drone is descending for} \ \ \dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)