A man on a railroad platform observed that a train passed by the point where he was standing in 10 seconds, and that the same train passed completely through a station, which is 308 meters long, in 24 seconds. How long was the train and how fast was it going? Thanks for help.
+129840 A man on a railroad platform observed that a train passed by the point where he was standing in 10 seconds, and that the same train passed completely through a station, which is 308 meters long, in 24 seconds. How long was the train and how fast was it going? Thanks for help.
Let L be the length of the train [ in meters ]
Then......in the first situation......it's rate can be modeled by : Distance traveled / time =
rate = L /10
In the second situation.....if we consider the back of the train, it travels [ L + 308] m in 24 seconds = rate
So.......equating rates, we have that
L / 10 = [ L + 308] / 24 cross multiply
24 L = 10 [ L + 308} simplify
24 L = 10L + 3080
14L = 3080
L = 220 m
And we can use L/ 10 to find its rate = 220 m / 10 s = 22 m/s
:
+118654 A man on a railroad platform observed that a train passed by the point where he was standing in 10 seconds, and that the same train passed completely through a station, which is 308 meters long, in 24 seconds. How long was the train and how fast was it going? Thanks for help.
I drew a pic to help me think about it.
\(speed=\frac{308m}{14sec}\\=\frac{308*60*60}{14*1000}=79.2km/hour\\~\\ \text{length of train}=\frac{308m}{14sec}\times 10sec=220m\)
+14985 A man on a railroad platform observed that a train passed by the point where he was standing in 10 seconds, and that the same train passed completely through a station, which is 308 meters long, in 24 seconds. How long was the train and how fast was it going?
\(train \ l_t\)
\(speed \ v\)
\(station \ l_s=308m\)
\(platform-time \ t_p=10sec\)
\(station-time \ t_s=24sec\)
\(v\times t_p=l_t\)
\(v\times t_s=l_t+l_s\)
\({\large v=\frac{l_t}{t_p}=\frac{l_t+l_s}{t_s}}\)
\(\large \frac{l_t}{10sec}=\frac{l_t+308m}{24sec}\)
\(24l_t=10l_t+3080m\)
\(14l_t=3080m\)
\(\large v=\frac{l_t}{t_p}=\frac{220m}{10sec}\times\frac{km}{1000m}\times \frac {3600sec}{h}\)
!
Let the speed of the train=S,
Let the Length of the train =L, so we have:
L/10 = S, and
[308+ L]/24 =L/10, solve for L
L=220 meters long
S=22 m/s=speed of train=79.2 km/h
+129840 A man on a railroad platform observed that a train passed by the point where he was standing in 10 seconds, and that the same train passed completely through a station, which is 308 meters long, in 24 seconds. How long was the train and how fast was it going? Thanks for help.
Let L be the length of the train [ in meters ]
Then......in the first situation......it's rate can be modeled by : Distance traveled / time =
rate = L /10
In the second situation.....if we consider the back of the train, it travels [ L + 308] m in 24 seconds = rate
So.......equating rates, we have that
L / 10 = [ L + 308] / 24 cross multiply
24 L = 10 [ L + 308} simplify
24 L = 10L + 3080
14L = 3080
L = 220 m
And we can use L/ 10 to find its rate = 220 m / 10 s = 22 m/s
:
