Two cars, P and Q move toward each other, getting 8\ m closer every...

Let the speeds of car P and car Q be $v_P$ and $v_Q$ respectively, and $v_P > v_Q$.

When the cars move towards each other, their relative speed is the sum of their individual speeds. Since they get $8\ \text{m}$ closer every second:

$v_P + v_Q = 8$

When the cars move in the same direction, their relative speed is the difference of their individual speeds. Since they get $0.6\ \text{m}$ closer every second:

$v_P - v_Q = 0.6$

Adding the two equations, we get:

$2v_P = 8.6 \Rightarrow v_P = 4.3\ \text{m/s}$

Subtracting the second equation from the first, we get:

$2v_Q = 7.4 \Rightarrow v_Q = 3.7\ \text{m/s}$

The ratio of the speeds of car P to car Q is:

$\dfrac{v_P}{v_Q} = \dfrac{4.3}{3.7} = \dfrac{43}{37}$

Answer: $43:37$