I have a math problem from a math contest I will be taking soon that I simply cannot understand how they got their answer.
If Train A leaves at noon from San Francisco and heads for Chicago going 40 mph. Two hours later Train B leaves the same station, also for Chicago, traveling 60mph. How long until Train B overtakes Train A?
I got 6 P.M. but the answer key says 4 P.M.?
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$\begingroup$The time when the two trains will meet is going to be the solution to the following equation (the intersection of two straight lines) where $t\ge0$ and $t=0$ corresponds to $12:00$ PM (noon):
$$ 40t=60(t-2)\implies\\ 40t=60t-120\implies\\ t=6\ P.M. $$
But the second train (B) departed at $2$ P.M. Therefore, it's $6-2=4$ hours before the second train (B) catches up with the first one (A). I don't know why it says P.M. in the answer key, but the answer to the question "how long until train B overtakes train A" should be the number of hours because it's the difference between two points in time designated as $6$ P.M. and $2$ P.M and that should be measured just in hours.
$\endgroup$ $\begingroup$Train B starts with a lag of $2\cdot40$ km.
It can make up $60-40$ km deficit in an hour
So, it will require $80/20$ hours to meet train $A$
$\endgroup$ $\begingroup$Your calculation of 6pm as the time of passing is correct. In the two hours where only Train A is traveling, it goes a total of 80 miles (40mph x 2 hours). The relative velocity between the trains is 20mph (60mph - 40mph). So, it takes 4 hours for train B to close the 80 miles between the two trains (80 miles / 20mph). Since Train B left at 2pm, it will pass Train A at 6pm.
Train B takes 4 hours to pass Train A, and this occurs at 6pm. The question, as posted, asks how long it takes for Train B to pass Train A, so the answer is 4 hours, not 4pm or 6pm.
$\endgroup$ $\begingroup$Train A leaves at time $t_{A0} = 0$ traveling at $40$ mph. Train B leaves at time $t_{B0} = t_{A0} + 2 = 2$ traveling at $60$ mph. Since velocity is distance divided by time (look at the units), we can write
$$v=\dfrac{d}{t}$$
Rearranging this yields the distance as a function of time:
$$d(t)=vt$$
Therefore, train A will have a distance of
$$d_A(t)=v_At = 40t$$
Train B will have a distance of
$$d_B(t)=v_Bt=60t$$
For the first two hours, train A travels alone, meaning it covers a distance of
$$d_A(2) = (40mph)\cdot(2hrs)=80miles$$
Now, making the start of train B our starting point, train A has an initial distance of $80$ miles, so it's distance becomes
$$d_A(t) = 80 +40t$$
Now we need to find the time that they meet, meaning we need to see when their distances are equal
$$\begin{align}d_A(t) &= d_B(t) \\ 80+40t&=60t \\ 80&=20t \\ t&=4hours \end{align}$$
We see that their time is $4$ hours from the start of train B, or $6$ hours from the start of train A. This implies that they meet at $6:00$ P.M.
$\endgroup$ $\begingroup$By the time train $B$ overtakes train $A$ each will make distance $S$. Thus, "time travelled by train $B$" = S/60 = "time travelled by train $A - 2$ hours". In other words, $S/60 = S/40 -2$, so $S=240$ (miles). As train $B$ travels at $60$ mph it needs $4$ hours to overtake train $A$.
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