91Ó°ÊÓ

• Initial velocity = 22 m/s.
• Deceleration rate = 0.150 m/s²
• 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,

$$\begin{gathered} 210=\left(22\right)t+\frac{1}{2}(-0.15)t^2 \\ -0.075t^2+22t-210=0 \end{gathered}$$

Hence the root of the equations are:

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

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,

$$\begin{gathered} V=22+(-0.15)×(9.879) \\ V=20.518\text{m/s} \end{gathered}$$

Hence the velocity of the nose is20.518 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,

$$\begin{gathered} 340=\left(22\right)t+\frac{1}{2}(-0.15)t^2 \\ 0.075t^2-22t+340=0 \end{gathered}$$

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

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

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,

$$\begin{gathered} V=22+(-0.15)×(16.3) \\ V=19.5\text{m/s} \end{gathered}$$

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