Expert: Scott A Wilson - 10/28/2009

Question
Two cars left points A and B simultaneously, traveling towards each other. 9 hours after their meeting, the car traveling from A arrived at B, and 16 hours after their meeting, the car traveling from b arrived at A. How many hours did it take the slower car to cover the whole distance?

Answer
One of the basic physice equation is d = rt, where d=distance, r=rate, and t=time.

I will try to remember to include a graph and hopefully you won't have to write back for it.

d1=distance car from A travelled
d2=distance car from B travelled
rA=rate of car A
rB=rate of car B
t=time until they meet

To start out, they both get to some point at the same time.
The equations for this are d1 = t•rA and d2 = t•rB.

After meeting, car B took 16 more hours, so d1 = 16•rB.
After meeting, car A took 9 more hours, so d2 = 9•rA.

Take all four of these equation and put in d1/d2 = d1/d2 as follows: 16•rB/(9•rA) = t•rA/(t•rB).

Notet the first thing to do is cancel the t's, giving: 16•rB/(9•rA) = rA/rB.

Now cross multiply, giving: 16•rB² = 9•rA².

Take the squareroots of both sides, giving: 4•rB = 3•rA.

This says that car A is going 4/3 as fast as car B.

This means that it will take car (3/4)16 = 12 hours to start his leg and
car B (4/3)9 = 12 hours to start his leg.  Since they are the same, that
puts them both at that meeting point at the same time.

From this, since it can be seen that B is slower, it will take car B 9 + 16 = 25 hours total.