Question

a train starting from rest attains a velocity of 72 km per hour in 5 minute as human the acceleration is uniform find
1) the acceleration
2) the distance travelled by the train for attaining this velocity ​

Answer 1

ㅤ$\;\;\underline{\textbf{\textsf{Given:-}}}$

• Initial velocity , u = 0m/s

• Final velocity , v = 72km/h

(Converting velocity km/hr into m/s)

$\sf \dashrightarrow v = 72\times \dfrac{5}{18} ms^{-1} \\\\ \sf \dashrightarrow v = 4\times 5 ms^{-1} \\\\ \sf \dashrightarrow v = 20ms^{-1}$

• Time taken, t = 5min 60 sec or 300 sec

ㅤ$\;\;\underline{\textbf{\textsf{To Find :-}}}$

• Acceleration, a

• Distance travelled, s

ㅤ$\;\;\underline{\textbf{\textsf{Solution :-}}}$

$\underline{\:\textsf{ Using 1st equation of motion :}}$

$\sf \dashrightarrow v = u + at \\\\ \sf \dashrightarrow a = \dfrac{v-u}{t} \\\\ \sf \dashrightarrow a = \dfrac{20 - 0}{300} \\\\ \sf \dashrightarrow a = \dfrac{20}{300} \\\\ \sf \dashrightarrow a = \dfrac{1}{15} \\\\ \sf \dashrightarrow a = 0.066$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Acceleration is </p><p>\textbf{0.066 m/s². }}}.$

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Again,

$\underline{\:\textsf{ Using 2nd equation of motion :}}$

$\sf \dashrightarrow s = ut+\dfrac{1}{2}at^{2} \\\\ \sf \dashrightarrow s = 0\times 300 + \dfrac{1}{2} \times \dfrac{1}{15}\times 300\times 300 \\\\ \sf \dashrightarrow s = 0 + 10\times 300 \\\\ \sf \dashrightarrow s = 3000$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ The distance travelled is </p><p>\textbf{ 3000m or 3km}}}.$

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ㅤ$\;\;\underline{\textbf{\textsf{ Know More :-}}}$

$(\bf 3rd \: equation \: of \: motion)$

$\boxed{\sf v^{2} - u^{2} = 2as}$

Where,

›› s = Distance Covered

›› u = Initial Velocity

›› v = Final Velocity

›› a = Acceleration

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