Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

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 m/s²as it goes through. The station is 210m 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, when does the end of the train leave the station?
(d) What is the velocity of the end of the train as it leaves?

a) 9.879 s.

b) 20.518 m/s.

c) 16.3 s.

d) 19.5 m/s.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Given Data

Initial velocity = 22 m/s.
Deceleration rate = 0.150 m/s²
Length of the station = 210 m.

Time period of the train

Since the train is only translating in a straight line and experimenting with a constant deceleration throughout the station, whose length is 210 m.

The time required for the nose of the train to reach the end of the station can be found with the help of the following motion formula:

$d = u t + \frac{1}{2} a t^{2}$

Here d is the distance traveled, u is the intial velocity, a is the acceleration, and t is the time.

Substituting the values in the above expression, we get,

$210 = (22) t + \frac{1}{2} (- 0.15) t^{2} - 0.075 t^{2} + 22 t - 210 = 0$

Hence the root of the equations are:

$T_{1} = 283.455 s and T_{2} = 9.879 s$

Both methods are physically sound, though second roots better depict the train's braking process.

Hence the time the nose is at the station is $9.879 s$ .

Speed of the nose

This expression gives the speed at which the nose leaves the station:

V = U + at

Here V is thr final velocity, U is the intial velocity, a is the acceleration rate, and t is the time.

Substituting the values in the above expression, we get,

$V = 22 + (- 0.15) \times (9.879) V = 20.518 m/s$

Hence the velocity of the nose is 20.518 m/s.

The end of the train's departure from the station

If the train is 130 meters long, the end of the train's departure from the station is:

The total distance will be $210 + 130 = 340 m$ .

The time at the end of the train leaves the station can be calculated as:

$d = u t + \frac{1}{2} a t^{2}$

Here d is the distance traveled, u is the intial velocity, a is the acceleration, and t is the time.

Substituting the values in the above expression, we get,

$340 = (22) t + \frac{1}{2} (- 0.15) t^{2} 0.075 t^{2} - 22 t + 340 = 0$

By solving the equation, we get the root of t as :

$T_{1} = 16.3 s and T_{2} = 276.9 s$

Here time will be 16.3 s as it is more feasible.

The velocity of the train's terminus as it leaves is calculated as follows:

V = U +at

Here V is thr final velocity, U is the intial velocity, a is the acceleration rate, and t is the time.

Substituting the values in the above expression, we get,

$V = 22 + (- 0.15) \times (16.3) V = 19.5 m/s$

Hence the speed of the end of the train will be 19.5 m/s .

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Given Data

Time period of the train

Speed of the nose

The end of the train's departure from the station

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College Physics (Urone)

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Time period of the train

Speed of the nose

The end of the train's departure from the station

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