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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?

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Ronnie, you're using the word "said" quite often in your questions, in surroundings where it is a rather non-standard choice of word. Usually "said X" is a lawyery/formal way of saying "the X that I spoke about previously", but in your question you seem to mean "some given X", "a known X" or "a certain X", or occasionally "an unknown X". Is your first language perhaps not English? –  Henning Makholm Sep 14 '11 at 23:57
    
Haha i learned the use said in that context from a tv show and i've been using it like that ever since. sorry for the confusion –  Ronnie.j Sep 15 '11 at 0:00
    
like i would say "but said person (to hide identity) said..., so according to said question..., etc. –  Ronnie.j Sep 15 '11 at 0:03
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Question: What do Archimedes, Euler, Gauss have in common? Answer: They did not watch TV. –  André Nicolas Sep 15 '11 at 0:37
    
that was my second guess –  Ronnie.j Sep 15 '11 at 1:14

3 Answers 3

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.

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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? –  Ronnie.j Sep 15 '11 at 0:07
    
Exactly. They're closing at 95 miles per hour, and they need to close 150 miles. –  David Schwartz Sep 15 '11 at 5:51

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.

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(+1) I think your solution is the faster one. –  Américo Tavares Sep 15 '11 at 12:11

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

enter image description here

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

$$t=\frac{d}{v_A+v_B}=\frac{150}{90+5}=\frac{30}{19}\approx 1.5789\text{ hours}.$$

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