# Four cars A,B,C,D are moving at constant speeds on the same road

I need help solving the following problem:

Four cars A,B,C,D are moving at constant speeds on the same road (not necessarily in the same direction). Car A passed car B at 8 a.m. and car C at 9 a.m. Car A met car D at 10 a.m. Car D met car B at 12 p.m. and car C at 2 p.m. What time did car B pass car C?

Clarifications: all times occur in one day

What I have worked out so far:

I have determined that Car A,B, and C are traveling in one direction, while car D is traveling in the other direction.

Car B must have passed car C in between the hours of 10 a.m. and 12 p.m.

I think I need to find the speed of the cars and then work from there, but I am not sure.

• I don't see why C has to be moving with A and B. It could be moving the same way D is, I think. Even if C is moving the same direction as A and B, B and C could meet as early as 9 – Ross Millikan Sep 11 '18 at 5:18
• Everything could move in the same direction if you choose a reference frame that moves faster than everything in the other direction. Thanks to Galilean relativity this will not change any of the crossing times. It might simplify calculations to measure everything in a frame where the speed of $A$ is $0$. – Henning Makholm Sep 11 '18 at 11:07

I think that you are paying too much attention to travelling directions. Just make things simple:

• Suppose that all cars are travelling in the same direction along the $x$-axis with their coordinates increasing. Denote their velocities with $v_A$, $v_B$, $v_C$, $v_D$
• Time $t$ starts at 8AM, when cars A and B meet
• The origin of $x$ is where cars A and B meet.
• Positions of cars C and D at $t=0$ are unknown. Denote that positions with $d_C$ and $d_D$.

You have the following set of equations that you have to solve for $t$:

$$v_A \cdot 1 = v_C \cdot 1+d_C\tag{1}$$

$$v_A \cdot 2 = v_D \cdot 2+d_D\tag{2}$$

$$v_D \cdot 4 + d_D = v_B \cdot 4\tag{3}$$

$$v_D \cdot 6 + d_D = v_C \cdot 6 + d_C\tag{4}$$

$$v_B \cdot t = v_C \cdot t + d_C\tag{5}$$

Eliminate $v_A$ from (1) and (2) and you get:

$$2v_C+2d_C = 2v_D+d_D\tag{6}$$

From (3) and (6) express $v_D$ and $d_D$ in terms of $v_B$, $v_C$ and $d_C$:

$$v_D=2v_B-v_C-d_C\tag{7}$$

$$d_D=4v_C+4d_C-4v_B\tag{8}$$

Now replace (7) and (8) into (4) and you get:

$$8v_B-8v_C=3d_C\tag{9}\implies \frac{d_C}{v_B-v_C}=\frac{8}{3}$$

From (5) and (9) it's now obvious that:

$$t=\frac{d_C}{v_B-v_C}=\frac83$$

..which is 2 hours and 40 minutes (after 8AM). So cars B and C will meet at 10:40AM.

It is given that the four cars $A, B, C, D$ drive at constant speed. Therefore we can write down expressions linear in time $t$ for their (instantaneous) positions $X$:

$$X(A) = V(A) * t$$

$$X(B) = V(B) * t + P(B)$$

$$X(C) = V(C) * t + P(C)$$

$$X(D) = V(D) * t + P(D)$$

For convenience we have set the initial position of car $A$ at $0$. We thus have a model with $7$ parameters. Now we use the information on car $A$ passing the other three cars. This enables us to determine the initial positions of cars $B, C, D$. They are given by:

$$P(B) = (V(A) - V(B))* 8$$

$$P(C) = (V(A) - V(C)) * 9$$

$$P(D) = (V(A) - V(D)) * 10$$

Next we use the information on car $D$ passing cars $B$ and $C$. This yields:

$$2* V(D) + 2*V(A) = 4*V(B)$$

$$4*V(D) + V(A)= 5*V(C)$$

We combine these two results in order to eliminate $V(D)$. This gives us:

$$3*V(A) = 8*V(B) - 5*V(C)$$

Substituting this result into the formulas for the positions of car $B$ and $C$, we determine that their positions are equal at $t = 32/3$, which corresponds to 40 minutes past 10 (a.m.).

• I believe this solution is correct. I got the same result, with the same strategy, but taking B as a fixed point and letting 8am be the initial time. I believe that taking B as the point of reference makes the solution a tiny bit simpler. – Barbosa Sep 11 '18 at 7:45