Question

2. A train starting from a railway station attains a speed of 40 km/hr in 10 min. Find its acceleration.​

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \blue {\underline{\underline{Provided\: that:}}}$

$\to \tt {Initial\: velocity (u)=0}$

➻As the train is Starting from the rest position,initial velocity is considered as zero.

━═━═━═━═━═━═━═━═━═━$\to \tt {Final\: velocity (v)=40km/hr}$

➻We need to convert km/hr into m/s.

➻To perform such conversion,we need to multilply the given value with the fraction $\large \tt {\frac {5}{18}}$

$\large \to \tt {v=40×{\frac{5}{18}}}$

$\large \to \tt {v=\frac{200}{18}}$

$\to \tt \green {\fbox{Final\: velocity(v)=11.11m/s}}$

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$\to \tt {Time\: interval(t)=10minutes}$

➻we need to convert minutes into seconds

$\to \tt {1minute=60seconds}$

$\to \tt {10minutes=?seconds}$

$\to \tt {t=10×60seconds}$

$\to \tt \green {\fbox{Time\: interval(t)=600s}}$

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$\sf \pink {\underline{\underline{To\: determine:}}}$

$\to \tt {Acceleration\: of\: the\: train(a)=?}$

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$\sf \purple {\underline{\underline{Using\: 1^{st}\: kinematic\: equation}}}$

$\implies \tt {\fbox{v=u+at}}$

➻supplanting the given values,

$\implies \tt {11.11=0+600×a}$

$\implies \tt {11.11=600a}$

$\large \implies \tt {a=\frac{11.11}{600}}$

$\colon \implies \tt \green {\boxed{\underline{Acceleration (a)=0.0185m/s^{2}}}}$

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$\sf \orange {\underline{\underline{Henceforth,}}}$

➻The acceleration of the train is $\tt {\underline{0.0185m/s^{2}}}$respectively.

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Answer 2

Answer:

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