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

Question

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.

Thanks for your help in advance

Answer 1

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.

Answer 2

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.).