How to calculate time in this condition? A starts at 11:00AM and travels at a speed of 4km/hr. B starts at 1:00PM and travels at 1km/hr for the first 1hr and 2km/hr for the next hr and so on. At what time they will meet each other ?
Note - Both A and B starts from same point and travel in same direction
Possible answer is 8:45 PM. (This is possible answer, I don't know this is right or wrong)
How to solve this question and which method is best to solve this type of question if time interval of start up of two objects is long?
 A: Let $t$ be the amount of time, in hours, that B has travelled when B catches up to A.
Note that A has travelled a distance $(2+t)(4)=8+4t$. 
In $3$ hours B has travelled $1+2+4$, not enough. 
In $4$ hours B has travelled $1+2+4+8$, and A has travelled $24$, so B is still behind.
In $5$ hours B has travelled $1+2+4+8+16$, and A has travelled $28$, so B is ahead.
Thus it took more than $4$ hours but less than $5$.
During that time, B travelled $1+2+4+8 +16(t-4)$. Now solve the linear equation
$$8+4t=1+2+4+8+16(t-4).$$
Remark: With other much larger numbers, finding the greatest integer in the number of hours travelled by B could be unpleasant.
Added: TonyK in a comment interprets $1$, $2$, and so on as meaning $1,2,3,4,\dots$. This is perfectly reasonable, physically more reasonable than repeated doubling.  Let $n$ be the greatest integer $\le t$. In time $2+n$, A has travelled $8+4n$ and B has travelled $1+2+3+\cdots +n=\frac{n(n+1)}{2}$. We want 
$$\frac{n(n+1)}{2}\le 8+4n \lt \frac{(n+1)(n+2)}{2}.$$
Finding the right $n$ can be done by trial and error, but for this interpretation it can also be done algebraically. It turns out that $n=8$. Calculating as in the main answer, we find that in $8$ hours B has travelled a distance $36$, so we get
$$8+4t=36+(t-8)(9).$$ 
