Background
Let us assume we have a function, $f_{s}(\mathbf{x},\mathbf{v},t)$, which defines the number of particles of species $s$ in the following way:
$$
dN = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \ d^{3}x \ d^{3}v
$$
which tells us that $f_{s}(\mathbf{x},\mathbf{v},t)$ is the particle distribution function of species $s$ that defines a probability density in phase space. We can define moments of the distribution function as expectation values of any dynamical function, $g(\mathbf{x},\mathbf{v})$, as:
$$
\langle g\left( \mathbf{x}, \mathbf{v} \right) \rangle = \frac{ 1 }{ N } \int d^{3}x \ d^{3}v \ g\left( \mathbf{x}, \mathbf{v} \right) \ f\left( \mathbf{x}, \mathbf{v}, t \right)
$$
where $\langle Q \rangle$ is the ensemble average of quantity $Q$.
Application
If we define a set of fluid moments with similar format to that of central moments, then we have:
$$
\text{number density [$\# \ (unit \ volume)^{-1}$]: } n_{s} = \int d^{3}v \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{average or bulk velocity [$length \ (unit \ time)^{-1}$]: } \mathbf{U}_{s} = \frac{ 1 }{ n_{s} } \int d^{3}v \ \mathbf{v}\ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{kinetic energy density [$energy \ (unit \ volume)^{-1}$]: } W_{s} = \frac{ m_{s} }{ 2 } \int d^{3}v \ v^{2} \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{pressure tensor [$energy \ (unit \ volume)^{-1}$]: } \mathbb{P}_{s} = m_{s} \int d^{3}v \ \left( \mathbf{v} - \mathbf{U}_{s} \right) \left( \mathbf{v} - \mathbf{U}_{s} \right) \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{heat flux tensor [$energy \ flux \ (unit \ volume)^{-1}$]: } \left(\mathbb{Q}_{s}\right)_{i,j,k} = m_{s} \int d^{3}v \ \left( \mathbf{v} - \mathbf{U}_{s} \right)_{i} \left( \mathbf{v} - \mathbf{U}_{s} \right)_{j} \left( \mathbf{v} - \mathbf{U}_{s} \right)_{k} \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{etc.}
$$
where $m_{s}$ is the particle mass of species $s$, the product of $\mathbf{A} \mathbf{B}$ is a dyadic product, not to be confused with the dot product, and a flux is simply a quantity multiplied by a velocity (from just dimensional analysis and practical use in continuity equations).
In an ideal gas we can relate the pressure to the temperature through:
$$
\langle T_{s} \rangle = \frac{ 1 }{ 3 } Tr\left[ \frac{ \mathbb{P}_{s} }{ n_{s} k_{B} } \right]
$$
where $Tr\left[ \right]$ is the trace operator and $k_{B}$ is the Boltzmann constant. In a more general sense, the temperature can be (loosely) thought of as a sort of pseudotensor related to the pressure when normalized properly (i.e., by the density).
Answers
How can a Hot gas be Low Pressured?
If you look at the relationship between pressure and temperature I described above, then you can see that for low scalar values of $P_{s}$, even smaller values of $n_{s}$ can lead to large $T_{s}$. Thus, you can have a very hot, very tenuous gas that exerts effectively no pressure on a container. Remember, it's not just the speed of one collision, but the collective collisions of the particles that matters. If you gave a single particle the enough energy to impose the same effective momentum transfer on a wall as $10^{23}$ particles at much lower energies, it would not bounce off the wall but rather tear through it!
How can a High Pressured gas be Cold?
Similar to the previous answer, if we have large scalar values of $P_{s}$ and even larger values of $n_{s}$, then one can have small $T_{s}$. Again, from the previous answer I stated it is the collective effect of all the particles on the wall, not just the individual particles. So even though each particle may have a small kinetic energy, if you have $10^{23}$ hitting a wall all at once, the net effect can be large.
It depends on what temperature is. Two systems are in thermal equilibrium when the fractional change of their multiplicities $\Omega$ with energy $E$, $\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E}$, are equal to each other. Let us call this quantity $\beta$.
For a classical ideal gas of $N$ independent particles the number of accessible states $\Omega$ is proportional to the surface of a hypersphere in a phase space with $3N$ dimensions. The radius of that sphere is proportional to the square root of the kinetic energy $\sqrt{E}$, so that $\Omega(E) \propto E^\frac{3N-1}{2}.$
This is enough to see that for the ideal classical gas $\beta=\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E} = \frac{3N-1}{2} E^{-1}$ which is equal to $\frac{3N}{2} E^{-1} $ because $N$ is on the order of Avogadro's number.
From kinetic theory, the product $pV= \frac{2}{3}E$.
Combining these two expressions we find the equation of state of the ideal gas $$\beta pV = N.$$
Comparing this with the empirical ideal gas law we see that $\beta = \frac{1}{k_B T}.$
Best Answer
In principle, YES.
Remember, temperature is related to the velocity distribution of the particles inside the gas. The key word here is $collision$.
Accelerating the train and suddenly stopping it is akin to shaking a container that contains fluid once. The sudden acceleration and deceleration of the contained will impart momentum from the wall of the container to the particles inside. I see two cases.
(a) If the container contains liquid and is not completely full, you could increase its temperature because you are imparting momentum of the particles inside the liquid by virtue of them sloshing inside the container.
(b) If the container contains liquid and is completely full, you could increase its temperature in the same principle but not significantly. Similarly, for a gas, while in principle you could have the same effect as in a container filled with gas, the density of the gas is so much lower than the liquid, the effect would have to be negligible.