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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.

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Last updated date: 13th Jun 2024
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Answer
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Hint:In this problem, we are given with the acceleration of the train and the time taken by the train to cover some distance. We are also given that the train starts from rest hence, its initial velocity will be zero. We need to calculate the speed and the distance travelled during the given time period. Use kinematical equations to calculate the required quantities.

Complete step-by-step solution:We will assume that the train moves in a linear manner thus, the distance travelled will be equal to displacement and the speed of the train will equal to its velocity.
We are given that;
The train starts from rest therefore, its initial velocity will be zero, $u = 0\,m{s^{ - 1}}$
The train has a uniform acceleration of $0.2\,m{s^{ - 2}}$ , $a = 0.2\,m{s^{ - 2}}$
The time is given to be $5$ minutes, $t = 5 \times 60 = 300{s^{ - 1}}$
We need to find the speed acquired during this time $v$ , using kinematical equation we have
$v = u + at$
Substituting the given values, we have:
$v = 0 + 0.2 \times 300$
$ \Rightarrow v = 60\,m{s^{ - 1}}$
The speed during this time after $5$ minutes will be: $60\,m{s^{ - 1}}$
The distance $s$ travelled during this time will be given as:
$s = \dfrac{1}{2}a{t^2}$
Substituting the values, we have:
$s = \dfrac{1}{2} \times 0.2 \times {\left( {300} \right)^2}$
$ \Rightarrow s = 9000\,m$
The distance travelled during this time is $9000\,m$ or $9km$ .
Therefore, the speed of the train during the given time after $5$ minutes will be: $60\,m{s^{ - 1}}$ and the distance travelled during this time is $9000\,m$ or $9km$ .

Note:The train starts from rest and so the3 initial velocity of the train must be taken as zero. The time is given in minutes so the unit must be converted to seconds. This problem can be solved using graphical methods. The Acceleration-time graph is a constant curve and so the area of the velocity-time graph (triangle shaped) will be the displacement of the train.