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Solved Problem on Kinematics
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A high-speed train with a constant speed of 234 km/h runs through a 620 m long tunnel, the train's length is 160 m. What is the time interval to cross the tunnel?


Problem Data Problem diagram

The train has, a dimension essential to the problem, it is a large object.
It begins to cross the tunnel when the front of the train arrives at the tunnel entrance and ends when the rear of the train arrives at the tunnel exit (figure 1).

Picture 1 - 160 m long train entering a 620 m long tunnel; picture 2 - train leaving the tunnel; picture 3 - diagram with oriented trajectory, with a particle, starting from the origin and reaching a position 780 m ahead.
figure 1

We choose a frame of reference oriented to the right. The problem can be reduced to a particle, which represents the rear of the train, at the origin of the frame (S0=0) with speed v=234 km/h, and a point given by the sum of train and tunnel lengths \( S=L+L_{\text{T}}=160+620=780\;\text{m} \) representing the tunnel exit.

Solution

First, we convert the speed of train given in kilometers per hour (km/h) to meters per second (m/s) used in the International System of Units (S.I.)
\[ v=234\;\frac{\cancel{\text{km}}}{\cancel{\text{h}}} \times \frac{1000\;\text{m}}{1\;\cancel{\text{km}}} \times \frac{1\;\cancel{\text{h}}}{3600\;\text{s}}=\frac{234}{3,6}\;\frac{\text{m}}{\text{s}}=65\;\text{m/s} \]
The particle travels with constant speed, writing the equation of displacement as a function of time, we have
\[ \bbox[#99CCFF,10px] {S=S_{0}+vt} \]
\[ 780=0+65\;t\\ 65\;t=780\\ t=\frac{780}{65} \]
\[ \bbox[#FFCCCC,10px] {t=12\;\text{s}} \]

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