Bug and the grass, when will it climb? If grass is 3 cm long and bug climbs 2 cm/h and grass grows 1 cm/h after how many hours will it climb? My teacher said that it will climb after 1.5 hours, how?Bug will climb 2 cm, there will be 1 cm left and also 1 cm more because grass grows 1 cm/h so there will be 2 cm left for the bug to climb, it will climb 2 more and 1 will grow again, after that bug will finally climb up after 3 hours of climbing, what is wrong here?
 A: The difficulty is that the bug climbs at 2cm/h relative to the grass. Now how exactly does the grass grow, does every little bit of it expand, or does it put on new height at the top, or at the bottom?
You are assuming it puts on new height at the top. In which case your answer is correct. Your teacher is assuming it puts on new height at the bottom, so after 1.5 hours it has reached the top of the original stalk, which has now been got new stalk length 1.5cm underneath it.
The interesting case is the middle case. Maybe you would like to tackle that?
Oh, what does grass really do? Well you cannot expect a mathematician to be interested in that, can you? :)
[Added later] Well. It seems that the original question maybe had the bug climbing down! So Harald's reference is good evidence that the teacher is wrong. Teachers after all are expected to be almost omniscient :)
A: The height of the grass at time $t\geq0$ is given by
$$h(t)=3+t\ .$$
A grass molecule which is at height $y_0\in[0,3]$ at time $0$ moves proportionally  upwards according to $$y(t)={3+t\over 3} y_0\ .$$
It follows that the upwards speed of this molecule is given by
$$\dot y(t)={1\over3} y_0={y(t)\over 3+t}\ .$$
This means that when the bug is at height $b$ at time $t$ it gets a free ride upwards of speed ${b\over 3+t}$. The initial problem describing the movement of the bug is therefore given by
$$\dot b={b\over 3+t}+2\quad (t\geq0),\qquad b(0)=0\ .\tag{1}$$
The solution of $(1)$ is
$$b(t)=2(3+t)\log\left(1+{t\over3}\right)\ .$$
We now have to determine the $t$ for which $b(t)=3+t$. The solution is
$$t_*=3\bigl(\sqrt{e}-1\bigr)\doteq1.946\ .$$
