Distance and time So basically I have a said distance.  There is one object moving at said speed from one side to the other(lets call it A to B), and another object moving at a different speed(faster speed) from the other side (B to A). How would I find the time it would take for the second object to reach the first?
 A: Suppose $A$ and $B$ are originally $d$ distance apart. Let $r_A$ and $r_B$ be the speed of $A$ and $B$, respectively. Then $r_A t + r_B t = d$ where $t$ is the time it takes to meet up. So $t = \frac{d}{r_A + r_B}$. 
Intuitively, you think of two object moving toward each other. They meet up when the sum of the distance travel equal the entire original distance between. 
A: Reduce the problem to one involving only one speed.  Thus A and B move toward each other, and A moves at one speed and B at another, so how fast are they moving toward each other?  That's just the sum of the two speeds.  That's the speed involved.  So you know the speed.  And you know the distance.  And you want the time.  Distance equals rate times time.
A: Added: I use the same approach of this answer of mine to your question "Physics: Uniform Motion". 

Would a picture help? Here is a graphical description of the problem

From the statement of the problem we conclude that both objects have an uniform motion, one from A to B with speed $v_A$ and the other from B to A with speed $v_B$. At the instant $t$ the distance to $A$ of the object moving from A to B is given by
$$s_A=v_A t,\tag{1}$$
while the distance to $A$ of the other is 
$$s_B=d-v_Bt.\tag{2}$$ 
At $t=0$ this second object is at B, which is at a distance from A I call $d$. Since it moves towards A at constant speed $s_B$, if we subtract the distance $v_B t$ from $d$ we obtain how distant it is from A at the instant $t$. 
They cross each other when $$s_A=s_B\tag{3}.$$
If we insert $s_A$ given by $(1)$ and $s_B$ given by $(2)$ into $(3)$ and solve for $t$, we find the answer:
$$v_A t=d-v_Bt\Leftrightarrow t=\frac{d}{v_A+v_B}\tag{4}$$

so if A is moving 90 mph and B is moving 5mph on a road that is 150 miles long, it will take 1.579 hrs for them to meet, right? 

Yes. Using $(4)$, with $d=150$ miles, $v_A=90$ mph, $v_B=5$ mph, yields
$$t=\frac{d}{v_A+v_B}=\frac{150}{90+5}=\frac{30}{19}\approx 1.5789\text{ hours}.$$
