Question

14. A train starts from a station and moves with a constant acceleration for 2 minutes. If it covers a distance of 400 m in this period, find the acceleration.​

Answer 1

Answer:

Step-by-step explanation:

Explanation: Given details: Distance covered by the train (S)= 400 m. Time taken for the travel (t) = 2 minutes = 2 * 60 seconds = 120 seconds.

Answer 2

★Concept : we are provided that, a train has just started from the station which means its initial velocity is 0 m/s and it has traveled with constant acceleration for 2 minutes i.e 120 seconds. Also, in that particular interval, it has crossed a distance of 400 m. We are asked to calculate the value of that constant acceleration. So we just have to substitute the given values in the 2nd kinematical equation of motion to find the required answer.

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Given:-

➢ Initial velocity (u) = 0 m/s

➢ Time (t) = 2 minutes = 120 seconds

➢ Distance (s) = 400 m

By using the 2nd Kinematical equation,

$\\\quad\leadsto\quad\underline{\boxed{\sf S=ut+\dfrac{1}{2}at^{2} }}\bigstar$

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Calculation:-

$\\\quad\longrightarrow\quad \sf S = ut + \dfrac{1}{2}at^{2}$

$\\\quad\longrightarrow\quad\sf 400 = (0)(120)+\dfrac{1}{2} (a)(120)^{2}$

$\\\quad\longrightarrow\quad\sf 400 = \dfrac{1}{2}\times14400\times a$

$\\\quad\longrightarrow\quad\sf 400 = 7200\times a$

$\\\quad\longrightarrow\quad\sf a = \dfrac{400}{7200}$

$\\\quad\therefore\quad\boxed{\bf a = 0.55 m/s^{2} }\bigstar$

That's the required Answer!!

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