Question

starting from rest velocity is 5 m/s in 20 sec after some time the velocity decreases to 3 m/s in next 6 s calculate the acceleration in both the cases​

Answer 1

Answer:-

In first case

• Initial velocity ,u = 0m/s
• Final velocity ,v = 5m/s
• Time taken ,t = 20s

As we know that acceleration is defined as the rate of change in velocity.

• a = v-u/t

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

$:\implies$ a = 5-0/20

$:\implies$ a = 5/20

$:\implies$ a = 1/4

$:\implies$ a = 0.25m/s²

Hence, the acceleration in 1st case is 0.25m/s²

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Now, calculating the acceleration in the 2nd case

• Initial velocity ,u =5m/s
• Final velocity ,v = 3m/s
• Time taken ,t = 6

$:\implies$ a = v-u/t

$:\implies$ a = 3-5/6

$:\implies$ a = -2/6

$:\implies$ a = -1/3

$:\implies$ a = -0.33 m/s²

Here, negative sign show retardation

• Hence, the acceleration in 2nd case is 0.33m/s²

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Answer 2

Given that , Starting from rest velocity is 5 m/s in 20 sec after some time the velocity decreases to 3 m/s in next 6 s.

Exigency To Find : The Acceleration in both cases ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar \:\: \bf Acceleration \: : \: \sf The \: rate \:of \: of \: change \:of\: velocity\:is \:known \: as \: Acceleration \:. \:\\\\ \qquad\maltese\:\:\bf Formula \:for \: Acceleration\::$

$\qquad \dag\:\:\bigg\lgroup \pmb{\frak{ Acceleration \:(\: a\:)\:: \dfrac { v - u}{t} }}\bigg\rgroup$

⠀⠀⠀Here , v is the final velocity , u is the Initial velocity , t is the total time taken and a is the Acceleration.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Acceleration in first case ,

⠀⠀⠀For First case given that ,

• The Initial velocity ( u ) is 0 m/s [ as , it starts from rests ] .
• The Final Velocity ( v ) is 5 m/s .
• The Total Time taken ( t ) is 20 seconds .

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:v - u \;}{t}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:v - u \;}{t}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:5 - 0 \;}{20}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:5 \;}{20}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: 0.25 \:$

$\qquad \therefore \:\:\pmb{\underline{\purple{\frak{ Acceleration \:(\:or\:a\:)\:=\: 0.25\:\:m/s^2 }}} }\:\:\bigstar$

• Acceleration in first case is 0.25 m/s²

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Acceleration in Second case ,

⠀⠀⠀For Second case given that ,

• The Initial velocity ( u ) is 5 m/s [ as , it starts from rests ] .
• The Final Velocity ( v ) is 3 m/s .
• The Total Time taken ( t ) is 6 seconds .

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:v - u \;}{t}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:v - u \;}{t}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:3 - 5 \;}{6}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \dfrac{\:-2 \;}{6}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: \cancel {\dfrac{\:-2 \;}{6}}$

$\qquad \dashrightarrow \sf Acceleration \:(\:or\:a\:)\:=\: -0.33$

$\qquad \therefore \:\:\pmb{\underline{\purple{\frak{ Acceleration \:(\:or\:a\:)\:=\: -0.33\:\:m/s^2 }}} }\:\:\bigstar$

• Acceleration in Second case is -0.33 m/s²

$\qquad \therefore \:\:\underline {\sf \:Hence, \:Acceleration \:in \:First \:and \: Second \:case \: are \: \bf 0.25 \: ms^2 \: and \: -0.33 \:m/s^2 \:\:\sf , respectively\:.}$