algebra precalculus - Distance, time and speed

Train $A$ leaves from station P to station Q, and at the same time train $B$ starts from station Q to station P on a parallel track. After meeting at point R, train $A$ takes $4$ hours and train $B$ takes $2$ hours to reach their respective destinations. Approximately, by what % is speed of train $B$ more than that of train $A$?

Answer

Assuming constant speed :

P --------- R --------------------------- Q

A --------> <--------------------------- B

$v_a = RQ / 4$ (in km per hour for example)

$v_b = PR / 2$

But you also have that they covered the other distance in the same amount of time $\tau$ :

$v_a = PR / \tau$

$v_a = RQ / \tau$

Now if you divide $v_a$ with $v_b$ on the 2 different cases :

$v_a/v_b = PR / RQ$

$v_a/v_b = RQ / 2 PR$

This means $RQ^2 = 2 PR^2$. If you take the square root, you have that :

$$v_a/v_b = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

As for the correct answer in approximated %, you have :

$v_b = \sqrt{2} v_a$

Therefore, the train B goes approximately 41% faster than the train A (1.41 = 100% + 41%)