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Ok so I have this homework problem. I dont want to give out the information because i want to put in the values myself, but basically I have one object moving at said speed. Said time later, another object leaves the same location as the object 1 and said speed faster. At some time later, object 2 is only said distance away from object 1. I have to find both their speeds. If this is too vague then i can re-write the question. Anybody got any tips?

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You should move this question to – Curiosity Sep 13 '11 at 23:27
ok thanks for the commentary – Ronnie.j Sep 13 '11 at 23:33
Your post does not make absolutely clear what information is given. Let $v$ be the speed of car 1. Let $v+k$ be the speed of car 2. Let $t_2$ be the time car 2 waits before taking off, and let $t$ be time it has travelled. Let $d$ be the distance it is still "behind." When car 2 started, it was $vt_2$ behind. It has gained $kt$, so it is $vt_2-kt$ behind. So $vt_2-kt=d$. If we know $k$, $t_2$, $t$, $d$, we can find $v$. [But if distance only is given, maybe car 2 is now ahead, then $kt-vt_2=d$. Whether this should be looked at depends on wording details of the question.] – André Nicolas Sep 13 '11 at 23:51
@Ronnie: It would have been better if you had given the problem fully, with all numbers, and requested that people not solve the problem for you, only give some idea of how to proceed. – André Nicolas Sep 14 '11 at 0:11
Yeah, let's credit Ronnie for an ingenious way of making sure that he will only get hints (ok, may be a formula) as opposed to a solution. Wouldn't mind seeing more of that actually! – Jyrki Lahtonen Sep 14 '11 at 6:55


A body is said to be in uniform motion when it travels equal distances in equal intervals of time (i.e. at a constant speed).

Wikipedia link.

enter image description here

My comment above. Distance traveled by the first object moving at velocity $v$ and departing at time $t=0$


Distance traveled by the second object moving at velocity $v+c$ and departing with a delay $t_{0}$ with respect to the first object

$$s_{2}=(v+c)(t-t_{0})\qquad t\geq t_{0}.\tag{2}$$

What is the distance $d$ between both objects at instant $t=t_{0}+t_{1}$ ($t_{1}$ after $t_{0}$)?


Added. Using information provided by OP in the comments:

A train leaves chicago at 12. 2 [$t_{0}$] hours later another train leaves chicago at a speed 50 [$c$] miles an hour faster than train A. After an hour [$t_{1}$], train B is 10 [$d$] miles behind train A. what are both their speeds [$v$ and $v+c=v+50$]?

You have two options.

(1) Replace the given numerical values in equations $(1)$ to $(3)$, and solve for $v$.

(2) Combine equations $(1)$ to $(3)$ to obtain the equation $d=vt_{0}-ct_{1}$, use the numerical data and solve for $v$.

If I used your [André Nicolas'] formula correctly, I got 30 miles for train 1 and 80 for train 2.

Let's use option (1) with $t_{0}=2$ h (after 12h), $c=50$ mph, $t_{1}=1$ h (after 12+2=14h), $d=10$ miles.

$$s_{1}=vt \text{ miles}.$$

$$s_{2}=(v+50)(t-2)\text{ miles}\qquad t\geq 2\text{ hours}.$$

At $t=t_{0}+t_{1}=2+1=3\text{ hours}$, we have

$$s_{1}=v\times 3=3v\text{ miles},$$

$$s_{2}=(v+50)(3-2)=v+50\text{ miles}$$


$$d=10=s_{1}-s_{2}=3v-\left( v+50\right) =2v-50\text{ miles},$$


$$10=2v-50\text{ miles},$$

whose solution is $v=30$ mph. So $v+c=v+50=30+50=80$ mph.

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You seem to be told the "said speeds" and then have to work them out. But in questions like these you could look at questions like:

  • How far away is the first when the second sets off?

  • What is the change in distance between them?

  • How much time does this change in distance take?

  • How fast does the distance between them change?

  • What is the difference between their speeds?

Added after comment:

  • You know how far apart they are after a known time

  • You know the difference in their speeds

  • So you can work out how far apart they were when the second car set off

  • This is how far the first car travelled in a known time

  • So you can work out how fast the first car travels

  • So you can work out how fast the second car travels

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Ok, well 1 (im numering your bullets) is not stated. 2. 10 miles. 3.1 hr. 4. isn't that the same as 3? and 4. object 2 is 50 mph faster than object 1 – Ronnie.j Sep 13 '11 at 23:45
My original equation was to try 3/x + 1/x+50=10. But that is incredibly wrong and I don't know what to do from there – Ronnie.j Sep 13 '11 at 23:50
If the change in distance between them is 10 miles and this takes 1 hour, then I would have thought their difference in speeds was 10 mph – Henry Sep 13 '11 at 23:54
haha i wish it was that easy xD. Ok let me rewrite this. Car1 leaves point A at said time. Some time later, Car2 leaves point A at a speed faster than car1. After some time from car 2 departure, car2 is only said distance away from car1. I need to find both their speeds – Ronnie.j Sep 13 '11 at 23:58
Thanks! with this and the help of others I was able to derive (i think) to the answer. much appreciated – Ronnie.j Sep 14 '11 at 0:49

The problem, in detail, is as follows.

A train leaves Chicago at 12. Two hours later, another train leaves Chicago at a speed 50 miles an hour faster than train A. After it has been travelling for 1 hour, train B is 10 miles behind train A. What are both their speeds?

We look at a slightly more general problem. Train A leaves a station, and travels at constant speed. Train B leaves the same station $h$ hours later, and travels in the same direction as Train A, at a speed of $k$ miles per hour more than Train A. After Train B has travelled for $t$ hours, Train B is $d$ miles behind Train A. Given $h$, $k$, $t$, and $d$, at what speed does Train A travel?

Solution: Let $v$ be the speed of Train A. When Train B leaves the station, Train A has been travelling for $h$ hours, at speed $v$. So when Train B leaves the station, Train A is $vh$ miles ahead of Train B.

After that, for every hour that elapses, the distance between the two trains decreases by $k$ miles. So after $t$ hours, the distance between the trains has decreased by $kt$ miles, and therefore it is
$$vh-kt.$$ But we were told that after Train B has been travelling for $t$ hours, Train B is $d$ miles behind Train A. It follows that $$vh-kt=d.$$ If we know any four of $v$, $h$, $k$, $t$, and $d$, we can use the above equation to solve for the fifth. In this problem, we know $h$, $k$, $t$, and $d$. We can solve the above equation explicitly for $v$. We get $vh=kt+d$, and therefore $$v=\frac{kt+d}{h}.$$

Another way: Start the clock at the instant that Train A leaves the station. After $h+t$ hours, Train A has travelled a distance $v(h+t)$. The speed of Train B is $v+k$. So $t$ hours after leaving the station, Train B has travelled a distance $(v+k)t$. But then Train B is $d$ miles behind Train A, so $$v(h+t) -(v+k)t=d.$$ Simplify the left-hand side. We get $$(vh+vt)-(vt+kt)=d,$$ and end up with $vh-kt=d$, the same equation as before.

Comments: $1$. The solution by Américo Tavares, which combines the algebraic with the graphical, is a much better one. It links, in a very clear way, the kinematic intuition with the geometric intuition. There is an awful lot that can be learned from it, in particular about the deep connection between velocities and slopes of certain lines. A thorough understanding of that solution is useful preparation for the calculus.

$2$. When you are writing up the problem, do not use the formula that I derived. Instead, use the idea, with the concrete numbers of the problem. Then everything will make physical sense, it will not be simply "a bunch of symbols."

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