Math Problem Statement

An express train passes through a station that is 204 m long. It enters with an initial speed of 23 m/s and decelerates at a rate of 0.15 m/s² as it goes through. (a) During what time interval is the front of the train within the station? (b) How fast is the train moving when its front leaves the station? (c) If the train is 124 m long, how long, in seconds, after the front of the train enters the station does the end of the train leave the station?

Solution

For part (c) of the problem, where the train is 124 meters long and we want to find how long, in seconds, it takes from the time the front of the train enters the station until the end of the train leaves the station:

  1. Determine the final speed of the train as it leaves the station: From part (b), we know the final speed of the train, vv, is 21.60 m/s.

  2. Calculate the total distance the train travels while in the station: Since the station is 204 meters long, the entire train must pass through this 204 meters, plus an additional 124 meters (the length of the train), making the total distance D=204+124=328D = 204 + 124 = 328 meters.

  3. Calculate the time taken using the kinematic equation: Given the initial velocity u=23u = 23 m/s, the final velocity v=21.60v = 21.60 m/s, and the acceleration a=0.15a = -0.15 m/s2^2 (deceleration), we can use the equation: v=u+atv = u + at to find the time tt it takes for the train to travel the 328 meters.

    First, we solve for tt from v=u+atv = u + at: t=vuat = \frac{v - u}{a}

    Then, using the time tt, we calculate the distance using: D=ut+12at2D = ut + \frac{1}{2}at^2

    Rearranging to find tt when D=328D = 328 meters: D=ut+12at2    t=2D2utaD = ut + \frac{1}{2}at^2 \implies t = \sqrt{\frac{2D - 2ut}{a}}

Let's calculate these steps to find the required time.The solution to the equation gives two potential values for time: approximately 14.99 seconds and 291.67 seconds. The time of 291.67 seconds is not physically meaningful in this context as it would suggest an unrealistically long duration for the train to travel through the station, considering the speeds involved and typical train dynamics.

Therefore, the time it takes from when the front of the train enters the station until the end of the train leaves the station, for a train that is 124 meters long, is approximately 14.99 seconds.

Would you like to delve into any part of the calculation in more detail or have any other questions on this topic? Here are some related queries you might consider:

  1. How does the length of the station affect the total time the train takes to pass through?
  2. What would be the effect of changing the rate of deceleration on the total time?
  3. How would an increase in the initial speed of the train affect the outcomes?
  4. Can you calculate the average speed of the train as it passes through the station?
  5. What would be the implications if the train were accelerating instead of decelerating?

Tip: Always double-check units in your calculations to ensure consistency and correctness.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Deceleration
Linear Motion

Formulas

v = u + at
D = ut + (1/2)at²

Theorems

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Suitable Grade Level

Grades 10-12