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Solved: An express train passes through a station. It
Chapter 2, Problem 36(choose chapter or problem)
An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of \(0.150 \mathrm{~m} / \mathrm{s}^{2}\) as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
Questions & Answers
QUESTION:
An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of \(0.150 \mathrm{~m} / \mathrm{s}^{2}\) as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
ANSWER:Step 1 of 6
Given data:
An express train passes through a station. It enters with an initial velocity u of 22.0 m/s and decelerates at a rate of \(0.150\;{\rm{m/}}{{\rm{s}}^2}\) as it goes through. The station is 210 m long.
Step 2 of 6
(a)
The deceleration is regarded as negative acceleration.
In the given case, the train is decelerating at a constant rate, so the kinetic equations can be used to solve the given problem.
The time taken for the nose of the train to cover the length of the platform can be determined as,
\(s = {v_0}t + \frac{1}{2}a{t^2}\)
Here, s is the distance covered, t is the time taken, a is the acceleration, and \({v_0}\) is the initial speed.
Since the train is decelerating, the acceleration must be taken as negative. So,
\(a = - 0.150\;{\rm{m/}}{{\rm{s}}^2}\)
Substitute the values in the above expression, and we get,
\(210 = 22t + \frac{1}{2}\left( { - 0.15} \right){t^2}\)
\(420 = 44t - 0.15{t^2}\)
Step 3 of 6
The above equati