What are the speeds of a bus and bike if they are heading in opposite directions? Unsure on how to go about solving for speed here Cities $A$ and $B$ are $70$ miles apart.  A biker leaves City $A$ at the same time that a bus leaves City $B$.  They travel toward each other and meet $84$ minutes after their departure at a point between $A$ and $B$.  The bus arrives at City $A$, stays there for $20$ minutes, and then heads back to City $B$.  The bus meets the biker again $2$ hours and $41$ minutes after their first meeting.  What are the speeds of the bus and the biker?
So far, I’ve created 2 variables $V_{\text{bus}}=$ bus speed and $V_{\text{bike}}=$ bike speed.  Where they meet for the first time, I’ve marked that as distance $C$.  From this I got $1.4 \text{ hrs} = C/{V_{\text{bike}}} = (70–C)/V_{\text{bus}}$.
To describe the time in which the bus begins heading back to City $B$, I did $(70/V_\text{bus})+1/3$ hrs which means the bike has traveled $V_\text{bike}((70/V_\text{bus})+1/3)$ by the time the bus leaves again.  This is all I have so far but I’m still unsure if I’m even heading in the right direction.
If it helps at all, the textbook I’m working out of has the answers as $35$ m/h and $15$ m/h but has no work.
 A: Let $A = (0,0)$ and $B = (70, 0) $, then the bike position is
$ p(t) = v_1 t $
while the bus position is
$ q(t) = 70 - v_2 t $
The time when they meet the first time is
$ t_1 = \dfrac{70}{v_1 + v_2} = 84 \text{ minutes } = \dfrac{7}{5} \text{ hours } $
Therefore, $v_1 + v_2 = 50 $
The bus reaches the origin at $t_1 = \dfrac{70}{v_2} $ , waits for $\dfrac{1}{3} $ hour, then heads back to $B$, therefore, for $ t \gt t_1 + \dfrac{1}{3} $  we have
$ p(t) = v_1 t $  and $ q(t) = v_2 (t - t_1 - \dfrac{1}{3 } ) $
These two are equal at $t = \dfrac{7}{5} + 2 \dfrac{41}{60} = \dfrac{245}{60} \text{hours} $
Hence,
$ v_1 \left( \dfrac{245}{60} \right) = v_2 ( \dfrac{245}{60} - \dfrac{20}{60} - \dfrac{70}{v_2} ) $
Simplifying,
$ 245 v_1 = 225 v_2 - 4200 $
From above, $v_2 = 50 - v_1$, hence
$ 245 v_1 = 225 (50 - v_1) - 4200 $
$\Rightarrow 470 v_1 = 7050 $
So $v_1 = \dfrac{7050}{470} = 15 \text{ mi / hour } $
From which it follows that $v_2 = 50 - v_1 = 35 \text{ mi / hour } $
A: Let the speeds of the biker and bus be $a,b$ mph respectively.
The first equation is easy, $(a+b) = \frac{70}{1.4} = 50\;\; ...\;\;[I]$
The next time they meet is $2h\, 41\, min = 161\, min$ later, but $B$ has traveled $20$ min less due to stop, and has still travelled $70$ miles more than $A$
Thus $b(84+161-20)/60 - a(84+161)/60= 70$
or $\frac{225}{60}b - \frac{245}{60}a = 70 ....[II]$
Solving, $a = 12\, mph,\; b = 15\, mph$

NOTE
This question already has an approved answer which I've upvoted, I made a mess at that time. Having found a simpler solution shorty thereafter, I hesitated to post, but now that it is all done and dusted, I thought a simpler solution would be useful to students appearing in competitive exams.
