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Two trains, each 400 meters long, pass each other completely in 10 seconds when they are moving in opposite directions. Moving in the same direction, they pass each other completely in 20 seconds.
Find the speed of the faster train.
40 meters per second
50 meters per second
60 meters per second
30 meters per second

Two trains, each $400$ meters long, pass each other completely in $10$ seconds when they are moving in opposite directions. Moving in the same direction, they pass each other completely in $20$ seconds. $\newline$ Find the speed of the faster train. $\newline$ $40$ meters per second $\newline$ $50$ meters per second $\newline$ $60$ meters per second $\newline$ $30$ meters per second

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Rate of travel: word problems

Full solution

Q. Two trains, each

400

meters long, pass each other completely in

10

seconds when they are moving in opposite directions. Moving in the same direction, they pass each other completely in

20

seconds.

\newline

Find the speed of the faster train.

\newline

40

meters per second

\newline

50

meters per second

\newline

60

meters per second

\newline

30

meters per second

Speed Calculation: Let's denote the speed of the first train as $v_1$ and the speed of the second train as $v_2$ . When the trains are moving in opposite directions, their relative speed is $v_1 + v_2$ . When they are moving in the same direction, their relative speed is $v_1 - v_2$ . $\newline$ The total distance covered when they pass each other is the sum of their lengths, which is $400 \text{ meters} + 400 \text{ meters} = 800 \text{ meters}$ . $\newline$ First, we calculate the relative speed when they are moving in opposite directions. $\newline$ Relative speed = Total distance / Time $\newline$ $v_1 + v_2 = 800 \text{ meters} / 10 \text{ seconds}$ $\newline$ $v_1 + v_2 = 80 \text{ meters/second}$
Opposite Directions: Next, we calculate the relative speed when they are moving in the same direction. $\newline$ Relative speed = Total distance / Time $\newline$ $v_1 - v_2 = 800 \text{ meters} / 20 \text{ seconds}$ $\newline$ $v_1 - v_2 = 40 \text{ meters/second}$
Same Direction: Now we have a system of two equations: $\newline$ $1$ ) $v_1 + v_2 = 80$ $\newline$ $2$ ) $v_1 - v_2 = 40$ $\newline$ We can solve this system by adding the two equations to eliminate $v_2$ . $\newline$ $(v_1 + v_2) + (v_1 - v_2) = 80 + 40$ $\newline$ $2v_1 = 120$ $\newline$ $v_1 = 120 / 2$ $\newline$ $v_1 = 60 \text{ meters/second}$

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