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Q2.5-36E

Expert-verifiedFound in: Page 85

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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/s ^{2 }as 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.

- Initial velocity = 22 m/s.
- Deceleration rate = 0.150 m/s
^{2} - Length of the station = 210 m.

a)

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:**

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

**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,

$210=\left(22\right)t+\frac{1}{2}(-0.15){t}^{2}\phantom{\rule{0ex}{0ex}}-0.075{t}^{2}+22t-210=0$

Hence the root of the equations are:

${\mathbf{T}}_{\mathbf{1}}=\mathbf{283}.\mathbf{455}\mathbf{s}\text{}\mathbf{and}\text{}{\mathbf{T}}_{\mathbf{2}}=\mathbf{9}.\mathbf{879}\text{s}$

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 $\mathbf{9}.\mathbf{879}\text{s}$ .

b)

**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,

$V=22+(-0.15)\times (9.879)\phantom{\rule{0ex}{0ex}}V=20.518\text{m/s}$

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

c)

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

The total distance will be $\mathbf{210}+\mathbf{130}=\mathbf{340}\mathbf{m}$.

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

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

**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,

$340=\left(22\right)t+\frac{1}{2}(-0.15){t}^{2}\phantom{\rule{0ex}{0ex}}0.075{t}^{2}-22t+340=0$

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

${\mathbf{T}}_{\mathbf{1}}=\mathbf{16}.\mathbf{3}\text{s}\mathbf{and}\text{}{\mathbf{T}}_{\mathbf{2}}=\mathbf{276}.\mathbf{9}\mathbf{s}$

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

d)

**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,

$V=22+(-0.15)\times (16.3)\phantom{\rule{0ex}{0ex}}V=19.5\text{m/s}$

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

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