Question

A train starting from rest moves with a uniform acceleration of \[0.2\,m{{s}^{-2}}\]for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

Answer 1

Hint: As the train is travelling with constant acceleration, its velocity will increase at a uniform rate. For constant acceleration, we can use the equations of motion to calculate the value of one variable and use it to calculate the other variable by substituting the corresponding values in it.

Formulae Used:
\[v=u+at\]
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]

Complete step by step solution:
Since the train is moving with uniform acceleration, we can use the equations of motion.
The equations of motion are applied to motion in a straight line or motion in one direction. Equations of motion give us a relationship between constant acceleration (\[a\]), initial velocity (\[u\]), final velocity (\[v\]), distance travelled (\[s\]) and time taken (\[t\]). They are-
\[v=u+at\] ………………………. (1)
\[{{v}^{2}}={{u}^{2}}+2as\] …………………….. (2)
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\] …………………... (3)
Using eq (1), we calculate speed as-
As initially the train was at rest so, \[u=0m{{s}^{-1}}\]
\[v=0+0.2\times (5\times 60s)\] [\[1\min =60s\]]
\[v=60m{{s}^{-1}}\]
Therefore, the train is moving with \[60m{{s}^{-1}}\] after 5 minutes. The distance travelled calculated from eq (3) will be,
\[\begin{align}
  & s=0+\dfrac{1}{2}\times 0.2\times {{(5\times 60)}^{2}} \\
 & s=9000m \\
\end{align}\]

In 5 minutes (\[5\times 60=300s\]), starting from rest, the train acquires a velocity of \[60m{{s}^{-1}}\] and travels a distance of \[9000m\].

Note:
Equations of motion are applied as long as acceleration is constant or no external forces are acting on the system. Convert units as required or it can give wrong answers, we prefer units to be in SI system because it is the most commonly followed system of units across the world.