With these questions about the motorcycle and the drone, it says that the distance between the motorcycle and drone is decreasing. If the distance is decreasing, then the rate of change of this distance would be negative, right? However, it seems that when I solve these problems if I put a negative sign in front of this rate of change (in this case 245 km/hr), it gives me the wrong answer. I have to input the 245 as a positive number. But that seems wrong, since it's a negative rate of change. Could someone explain? Thank you!
A drone is flying at 101 kilometers per hour at a constant altitude of 1 km above a straight road. The drone pilot uses radar to determine that an oncoming motorcycle is at a distance of exactly 2 kilometers from the drone, and that this distance is decreasing at 245 kilometers per hour.
Find the speed of the motorcycle (in kilometers per hour).
In your units, you wish to compare two right triangles, defined by the right projection of the drone on the ground, the drone, and the motorcyclist: at time 0, with sides 1 & $\sqrt{3}$ and hypotenuse 2; and at small time t (we'll see how small) with sides 1 & $\sqrt{3}-(101+v)t$ and hypotenuse 2-245t, where v is the speed of the motorcyclist.
But you want the base of the latter triangle, $\sqrt{3}-(101+v)t$, to be longer than the height, 1, as at the start, so take t to be small, and hence ignore terms of $O(t^2)$ in $$(\sqrt{3}-(101+v)t)^2 + 1 = (2-245t)^2. $$
The terms independent of t of course cancel, and ignoring the quadratic ones, the terms linear in t yield $$ 2\cdot 245= \sqrt{3} ~ (101+v) \leadsto \\ v\approx 182 . $$
Would you be unhappy with this sort of thing?