Two trains, each traveling at 60 kph [1 kilometer/minute] are on parallel tracks heading toward one another. One train is one-half kilometer long and the other train is two-thirds kilometer long. The fronts of the two trains pass the same point A at the same time. How many seconds does it take for the two trains to completely pass each other?
+26367 Two trains, each traveling at 60 kph [1 kilometer/minute] are on parallel tracks heading toward one another. One train is one-half kilometer long and the other train is two-thirds kilometer long. The fronts of the two trains pass the same point A at the same time. How many seconds does it take for the two trains to completely pass each other ?
$$v= 1\cdot \dfrac{km}{\text{min.}}\\\\
1.\ \text{ train speed*time: } \quad s_1 = vt \\
2.\ \text{ train speed*time: } \quad s_2 = vt \\\\
1.\ \text{ train long: } \quad l_1 = \frac{1}{2}\ km \\
2.\ \text{ train long: } \quad l_2 = \frac{2}{3}\ km \\\\
\boxed{s_1+s_2=l_1+l_2}\\\\
l_1+l_2 = vt+vt=2vt\\\\
\boxed{t=\dfrac{l_1+l_2}{2v}}\\\\
t = \dfrac{1}{2\cdot \dfrac{1\ km}{\ \text{min.}} }\cdot \left(
\dfrac{1}{2} + \dfrac{2}{3}
\right)\ km$$
$$\\t = \dfrac{1}{2}\cdot \dfrac{7}{6} \ \text{min.} \\\\
t = 0.58 \overline{3} \ \text{min.}\\ \\
t = \dfrac{1}{2}\cdot \dfrac{7}{6} * 60 \ s \\\\
t = \dfrac{7*10}{2}\ s = 7*5 \ s \\ \\
t = 35 \ s$$
+33615 The relative speed is 2 km/minute. That is, assume one of the trains is stationary and the other is travelling at the combined speed of both.
The total distance to be travelled before they completely pass each other is (1/2 + 2/3)km = 7/6 km.
Total time is therefore (7/6)/2 minutes = 7/12 minutes = 7*60/12 seconds = 7*5 seconds = 35 seconds
.
+118608 Check out this trian. It goes from Shanghai to the Shanghai airport.
It is the worlds fastest.
It doesn't actually run on the track it is a magnetic leviatation train.
It shows two trains passing each other too but you have to be watching LOL
I have travelled on this train, it was certainly an experience. :)))
+26367 Two trains, each traveling at 60 kph [1 kilometer/minute] are on parallel tracks heading toward one another. One train is one-half kilometer long and the other train is two-thirds kilometer long. The fronts of the two trains pass the same point A at the same time. How many seconds does it take for the two trains to completely pass each other ?
$$v= 1\cdot \dfrac{km}{\text{min.}}\\\\
1.\ \text{ train speed*time: } \quad s_1 = vt \\
2.\ \text{ train speed*time: } \quad s_2 = vt \\\\
1.\ \text{ train long: } \quad l_1 = \frac{1}{2}\ km \\
2.\ \text{ train long: } \quad l_2 = \frac{2}{3}\ km \\\\
\boxed{s_1+s_2=l_1+l_2}\\\\
l_1+l_2 = vt+vt=2vt\\\\
\boxed{t=\dfrac{l_1+l_2}{2v}}\\\\
t = \dfrac{1}{2\cdot \dfrac{1\ km}{\ \text{min.}} }\cdot \left(
\dfrac{1}{2} + \dfrac{2}{3}
\right)\ km$$
$$\\t = \dfrac{1}{2}\cdot \dfrac{7}{6} \ \text{min.} \\\\
t = 0.58 \overline{3} \ \text{min.}\\ \\
t = \dfrac{1}{2}\cdot \dfrac{7}{6} * 60 \ s \\\\
t = \dfrac{7*10}{2}\ s = 7*5 \ s \\ \\
t = 35 \ s$$
+33615 Nice picture Melody. This should really help people understand the problem and the solution. An example of a picture being worth a thousand words!
.
+118608 Thanks Alan. I hope it helps a bit anyway.
Did you wach that clip of the magnetic levitation train in Shanghai? The most amazing bit is when the two trains - each with many carriages pass. The passengers don't even 'see' the other train. It really is amazing!
You know, we use pictures and graphs to help illustrate things all he time but making sense of these 'aids' is a learned skill.
For anyone who is reading this I greatly encourage you to try and figure out the relevances of the pictorials that we show you. If you learn to interprete graphs etc properly it will help you enormously.
Never be afraid to ask questions. We actually enjoy being cross examinied. It is the only way that we really know we are teaching/helping people.
