A train starting from rest attains a velocity of 20m/s in 2 minutes. Assuming that the acceleration is uniform, the Acceleration will be [[1]] m/s² and the Distance traveled by the train while it attained its final velocity of 20m/s is [[1]] m.

We can apply the first equation of motion to get the answer to the first part of the question. Then apply the third equation of motion to find the distance traveled by the train: 

$v=u+a{t}^2$

${v}^2=u^2+2as$

a)The initial speed is u = 0 because the train starts from a stationary state. The terminal velocity of the train is given in the question, i.e. v = 20 m / s 

The time given in the question is 2 minutes = 120 seconds. 

 Therefore, applying the first equation of motion,

$v=u+at$

gives the value of acceleration.

Substituting a value into the above formula yields: 

$$\begin{gathered} v=u+at \\ \Rightarrow 20=0+a\times 120 \\ \Rightarrow a=\frac{1}{6}m/s^2 \end{gathered}$$

b)The initial speed is u = 0 because the train starts from a stationary state.

 The terminal velocity of the train is given in the question, i.e. v = 20 m / s 
The time given in the question is 2 minutes = 120 seconds.

Then apply the third equation of motion.$v^2=u^2+2as$, Where s is the distance.

$v^2=u^2+2as$

$$\begin{gathered} \Rightarrow {20}^2=0+2\times \frac{1}{6}\times s \\ \Rightarrow 400 = \frac{1}{3}\times s \\ \Rightarrow s=1200m \end{gathered}$$