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Answer:
Given that , A train moves from rest to a speed of 25 m/s in 30 seconds .
Exigency To Find : The magnitude of Acceleration of train ?
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⠀⠀⠀⠀⠀⠀⠀⠀⠀Given that ,
⠀⠀▪︎⠀The initial velocity ( u ) of train is 0 m/s [ as , train starts from rest ] .
⠀⠀▪︎⠀The final velocity ( v ) of train is 25 m/s .
⠀⠀▪︎⠀Total Time taken ( t ) is 30 seconds .
⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Magnitude of Acceleration of Train :
\begin{gathered}\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar \:\: \bf Acceleration \: : \: \sf The \: rate \:of \: of \: change \:of\: velocity\:is \:known \: as \: Acceleration \:. \:\\\\ \qquad\maltese\:\:\bf Formula \:for \: Acceleration\:: \\\\ \end{gathered}
†
As,Weknowthat:
As,Weknowthat:
★Acceleration:TherateofofchangeofvelocityisknownasAcceleration.
✠FormulaforAcceleration:
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \pmb{\frak{ Acceleration \:(\: a\:)\:: \dfrac { v - u}{t} }}\bigg\rgroup \\\\\end{gathered}
†
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⎪
⎪
⎪
⎧
Acceleration(a):
t
v−u
Acceleration(a):
t
v−u
⎭
⎪
⎪
⎪
⎫
⠀⠀⠀Here , v is the final velocity , u is the Initial velocity , t is the total time taken and a is the Acceleration.
\begin{gathered}\qquad \dashrightarrow \sf Acceleration \:(\: a\:)\:=\: \dfrac { v - u}{t} \:\:\\\\\qquad \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\qquad \dashrightarrow \sf Acceleration \:(\: a\:)\:=\: \dfrac { v - u}{t} \:\:\\\\\qquad \dashrightarrow \sf Acceleration \:(\: a\:)\:=\: \dfrac { 25 - 0}{3} \:\:\\\\\qquad \dashrightarrow \sf Acceleration \:(\: a\:)\:=\: \dfrac { 25 }{3} \:\:\\\\\qquad \dashrightarrow \sf Acceleration \:(\: a\:)\:=\: \cancel {\dfrac { 25 }{3}} \:\:\\\\\qquad \dashrightarrow \sf Acceleration \:(\: a\:)\:=\: \:0.833 \:\:\\\\\qquad \therefore \pmb{\underline{\purple{\frak{ \: Acceleration \:(\: or\:\:a\:)\:=\: \:0.833\:m/s^2 }} }}\bigstar \\\\\end{gathered}
⇢Acceleration(a)=
t
v−u
⋆NowBySubstitutingtheknownValues:
⇢Acceleration(a)=
t
v−u
⇢Acceleration(a)=
3
25−0
⇢Acceleration(a)=
3
25
⇢Acceleration(a)=
3
25
⇢Acceleration(a)=0.833
∴
Acceleration(ora)=0.833m/s
2
Acceleration(ora)=0.833m/s
2
★
\begin{gathered}\qquad \therefore \:\:\underline {\sf Hence, \:The\:Acceleration \:is \: \bf 0.833\:m/s^2 \:}\\\\\end{gathered}
∴
Hence,TheAccelerationis0.833m/s
question kya hai be
Answer:
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