Find time when 2 cars meet? There is a roadway between city A and B . A car P starts at 5:00 am from A and reaches B at 10:00 am. Another car Q starts from B at 7:00 am and reaches A at 9:00 am. Find the time when car P meets car Q ?
I did as follows
for car P , travelling from 5:00 am to 7:00 am cannot definitely meet car Q as Q  has not even started . 
Lets say at 7:00 am car P reached a point R, and lets assume total road way distance be D
so $D_{AR} = \frac{2\times D}{5} $
so lets see from 7:00 am to 9:00 am when they will meet . 
Let the cars P and Q are said to meet at a distance $D_{1}$ from R and $D_{1}$ from B.
The velocity ratio is inversely proportional to time ratio so it is $2/5$
$$ \frac{D_{1}}{D_{2}} = \frac{2}{5}$$
As distance travelled by car P between 7:00 to 9:00 (that is 2 hrs ) is $\frac{2\times D}{5} $
so the cars meet at $D_{AR} + \frac{2}{7}\times \frac{2\times D}{5} $ =  $ \frac{18 \times D }{35} $ from city A
SO Time is calculated as like 
To travel D distance if it takes 5 hours , then to travel $ \frac{18 \times D }{35} $ it should take 
$\displaystyle\frac{5}{D} \times \frac{18D}{35}$
So $\frac{18}{7}$ hours . Am  I right ? If wrong please correct . 
Even if my answer is correct please suggest if there are easier and logical ways to solve this sort of problems without equations . In the above also I have tried avoiding equations by using simple logic Distance proportional to velocity when speed is constant. 
 A: Let the meeting time be $t$, a real number with the whole portion being hours.  At time $t$, $P$ has covered $\frac {t-5}5$ of the distance.  $Q$ has covered $\frac {t-7}2$ of the distance.  These have to add to the whole distance, so $$\frac {t-5}5+\frac {t-7}2=1\\2t-10+5t-35=10\\7t=55\\t=7\frac67\approx 7:51:25.714$$
A: Let's say that cars velocity is constant.
Then car B travells $\frac{1}{2}$ of the path in one hour and car A travells $\frac{1}{5}$
of the path. We know that they move each to each other so when they meet the sum of the distances they've passed will be equals as the distance between place A and place B.
Then we'll have:
$$\frac{x}{2} + \frac{x+2}{5} = 1$$
$$\frac{5x + 4 + 2x}{10} = 1$$
$$7x + 4 = 10$$
$$7x = 6$$
$$x = \frac{6}{7}$$
It means that they'll meat $\frac{6}{7}$ hours after the second car have started travelling.
A: Normalize the distance to $1$.  Let $x$ be the location they meet at time $t$.  The distance traveled by P is $x=v_Pt, $ similarly for Q the distance traveled is $1-x=v_Q(t-2).$  $v_P$ and $v_Q$ will be $\frac15,$ and $\frac12$ respectively based on the total travel time.  Solve the two equations for $t$ and $x$.
