What is the average acceleration of a subway train that speeds up from $9.6\,m\,{s^{ - 1}}$ to $12\,m\,{s^{ - 1}}$ in $0.8\,s$ on a straight section of track?

Hint: Acceleration is defined as the change in the velocity per unit time.In the question we are given the initial and final velocity of the subway train and the time interval over which the velocity changes. We will simply plug in the known values in the formula and get the acceleration.

Formula used:
\[a = \dfrac{{v - u}}{t}\]
where $a$ is the acceleration, $v$ is the final velocity and $u$ is the initial velocity.

Complete step by step answer:
We know that acceleration is defined as the change in the velocity per unit time.Mathematically it is expressed as,
\[a = \dfrac{{\Delta v}}{t}\]
$ \Rightarrow \Delta v = v - u$
Substituting in the equation, we get
\[a = \dfrac{{\Delta v}}{t}\]
\[ \Rightarrow a = \dfrac{{v - u}}{t}\]
We are given that $u = 9.6\,m\,{s^{ - 1}}$ and $v = 12\,m\,{s^{ - 1}}$ . Also, the time in which the velocity changes is given to be $0.8\,s$ .
Now putting the known values in the above equation,
\[a = \dfrac{{12 - 9.6}}{{0.8}}\]
Further solving this equation, we get,
\[a = \dfrac{{2.4}}{{0.8}}\]
$\therefore a = 3\,m\,{s^{ - 2}}$

Therefore, the average acceleration of a subway train is $3\,m\,{s^{ - 2}}$.

Note: Since velocities are vector quantities, along with their magnitude, their direction also is very important. If in this question, the velocity would have changed its direction, then we would have considered $-v$ in place of $v$. When acceleration is constant, speed equations can also be applied. On rearranging the formula of average acceleration, we get the first speed equation.