Solved Problem on One-dimensional Motion
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Two trains of 120 and 280 meters in length move in straight parallel lines with constant speeds. When the two trains move in the same direction they are required 20 seconds for the first train to overtake the second, when they move in opposing directions are required 10 seconds for one to pass the other. Determine the speeds of the trains.


Problem data:
  • Length of train 1:    d1 = 120 m;
  • Length of train 2:    d2 = 280 m;
  • Time interval for overtaking in the same direction:    tA = 20 s;
  • Time interval for overtaking in opposite directions:    tB = 10 s.
Problem diagram:

If they move in the same direction the overtaking begins when, the front of the back train reaches the back of the front train, and ends when, the back of the first train passes from the front of the second train.

Figure 1

We choose a reference frame pointing to the right. The problem can be reduced to a point, which represents the back of the first train at the origin of the reference frame S01 = 0 with speed v1, and another point that represents the front of the second train, at a point given by the sum of the lengths of the two trains 120+280=400 m, ahead S02 = 400 m with v2 speed. Overtaking occurs when these two points meet (Figure 1).
If they move in opposite directions the overtaking begins when, the front of the back train meets the front of the front train, and ends when, the back of the first train passes by the back of the second train.

Figure 2

We choose a reference frame pointing to the right. The problem can be reduced to a point, which represents the back of the first train at the origin of the reference frame S01 = 0 with speed v1, and another point that represents the back of the second train at a point given by the sum of the lengths of the two trains 120+280=400 m, ahead S02 = 400 m with speed −v2. The overtaking occurs when these two points cross (Figure 2).

Solution

The equation of displacement as a function of time with constant velocity is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+vt} \tag{I} \end{gather} \]
writing the expression (I) for the two points in the same direction (Figure 1) for the first train
\[ \begin{gather} S_{1}=S_{01}+v_{1}t_{A}\\ S_{1}=0+v_{1}.20\\ S_{1}=20v_{1} \tag{II} \end{gather} \]
for the second train
\[ \begin{gather} S_{2}=S_{02}+v_{2}t_{A}\\ S_{2}=400+20v_{2} \tag{III} \end{gather} \]
Imposing the condition that when the two points meet they occupy the same position in the trajectory we equate the expressions (II) and (III)
\[ \begin{gather} S_{1}=S_{2}\\ 20v_{1}=400+20v_{2}\\ 20v_{1}-20v_{2}=400 \tag{IV} \end{gather} \]
Writing the expression of the two points for the motion in opposite directions (Figure 2) for the first train
\[ \begin{gather} S_{1}=S_{01}+v_{1}t_{B}\\ S_{1}=0+v_{1}.10\\ S_{1}=10v_{1} \tag{V} \end{gather} \]
for the second train
\[ \begin{gather} S_{2}=S_{02}-v_{2}t_{B}\\ S_{2}=400-10v_{2} \tag{VI} \end{gather} \]
Imposing the condition that when the two points cross, they occupy the same position in the trajectory we equate expressions (IV) and (V)
\[ \begin{gather} S_{1}=S_{2}\\ 10v_{1}=400-10v_{2}\\ 10v_{1}+10v_{2}=400 \tag{VII} \end{gather} \]
Expressions (IV) and (VII) can be written as a system of two equations to two variables, v1 e v2
\[ \left\{ \begin{array}{l} 20v_{1}-20v_{2}=400\\ 10v_{1}+10v_{2}=400 \end{array} \right. \tag{VIII} \]
multiplying the second equation of the system (VIII) by 2 and adding with the first equation
\[ \begin{gather} \qquad\qquad \left\{ \begin{array}{l} 20v_{1}-20v_{2}=400\\ 10v_{1}+10v_{2}=400 \qquad (\times 2) \end{array} \right.\\[10pt] \frac{ \begin{align} 20v_{1}-20v_{2}&=400\quad\;\\ (+)\;20v_{1}+20v_{2}&=800\quad\; \end{align} } {40v_{1}+0v_{2}=1200}\\ 40v_{1}=1200\\ v_{1}=\frac{1200}{40} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {v_{1}=30\;\text{m/s}} \]
Substituting this value v1 in any of the system equations (VIII), we have the value of v2, substituting in the second equation
\[ \begin{gather} 10\times 30+10v_{2}=400\\ 300+10v_{2}=400\\ 10v_{2}=400-300\\ 10v_{2}=100\\ v_{2}=\frac{1000}{10} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {v_{2}=10\;\text{m/s}} \]
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