Two cars are driving toward each other

algebra-precalculus

Two cars start at the same time and drive towards each other from towns A and B. When they first meet they noticed that they are 70 km away from A. When each of them arrived in the town which he travelled to, they immediately went back. This time they met 40 km away from B. Find the distance between the towns.

First I've tried to make a table before their first meeting and after that but I can't complete it. Can you give me some help and give me a joker? I would be very grateful!

Best Answer

The way to solve this sort of word problem is by algebra.

First, identify all the relevant unknowns. I find 5 (all distances are in km; for time, choose any timescale you like):

  • $d$ = the distance between the two towns (in km)
  • $v_A$ is the speed at which the man starting in town $A$ drives.
  • $v_B$ is the speed at which the man starting in town $B$ drives.
  • $t_1$ is the time between starting and their first meeting.
  • $t_2$ is the time between starting and their second meeting.

Then, figure out what relationships exist between the variables. I find 4:

  • At time $t_1$, man A has driven a distance of $70$. Man B has driven a distance of $(d - 70)$. Distance = speed times time, so $$v_A t_1 = 70\\v_Bt_1 = d - 70$$
  • At time $t_2$, each has driven the complete distance $d$ to the other town, and then back to the point $40$ km from $B$. So, recalling that man A begins his second trip from $B$ and man B from $A$, $$v_A t_2 = d + 40\\v_Bt_2 = d + (d - 40) = 2d - 40$$

Now, there is a problem here. I have 5 unknowns, but only 4 equations. That means that I cannot find a unique solution for all of the variables (you have to have as many equations as you have variables to get a single solution). However, it may still be possible to solve for particular variables, such as $d$.

One thing that stands out with those two pairs is that the 2nd equation in each subtracts a constant that is added in the first equation. So let's try adding the equations together. Adding the first pair of equations gives $$v_At_1 + v_Bt_1 = 70 + d - 70$$$$(v_A + v_B)t_1 = d \tag1$$ Adding the second pair of equations gives $$v_At_2 + v_Bt_2 = (d + 40) + (2d -40)$$$$(v_A + v_B)t_2 = 3d\tag2$$

And now substitute for $d$ from equation (1) into equation (2): $$(v_A + v_B) t_2 = 3\big((v_A + v_B)t_1\big)$$ If we divide both sides by $(v_A + v_B)$, we get $$t_2 = 3t_1$$ This allows us to eliminate $t_2$ from the original equations: $$v_A 3t_1 = d + 40\\v_B3t_1 = 2d - 40$$

Now we have $$v_A t_1 = 70\tag3$$ and $$3v_At_1 = d + 40 \tag4$$ Substituting for $v_A t_1$ from (3) into (4), we get $$3(70) = d + 40\\d = 170$$

The towns are $170$ km apart.

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