\star\underline{\mathtt\orange{⫷❥Q᭄} \mathfrak\blue{u~ }\mathfrak\blue{Σ} \mathbb\purple{ §}\mathtt\orange{T} \mathbb?\pink{iOn⫸}}\star\:⋆

⫷❥Q᭄u Σ§TiOn⫸

⋆

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?

{ { \underbrace{ \mathbb{ \red{GIVEN\ }}}}}

GiVeN

Initial \:velocity(u) =0m/sInitialvelocity(u)=0m/s

Final\:velocity (v) =8m/sFinalvelocity(v)=8m/s

Time\: taken (t) =4sTimetaken(t)=4s

{ { \underbrace{ \mathbb{ \red{To\:PrOvE\ }}}}}

ToPrOvE

Distance \:travelled \:of \:a \:particleDistancetravelledofaparticle

\star\underbrace{\mathtt\red{⫷❥ᴀ᭄} \mathtt\green{n~ }\mathtt\blue{ §} \mathtt\purple{₩}\mathtt\orange{Σ} \mathtt\pink{R⫸}}\star\:⋆

⫷❥ᴀ᭄n §₩ΣR⫸

⋆

{\boxed {\boxed {v=u+at}}}

v=u+at

v²=u²+2asv²=u²+2as

s=ut+\frac{1}{2} at²s=ut+

2

1

at²

v=Final\:velocityv=Finalvelocity

u=Initial\:velocityu=Initialvelocity

a=accelerationa=acceleration

t=Time\:takent=Timetaken

now\:substitute\:the \:valuesnowsubstitutethevalues

8=0+a(4)8=0+a(4)

8=a\times 48=a×4

a= \frac{8}{4}a=

4

8

{\boxed {\boxed {a=2m/s²}}}

a=2m/s²

________________________________

v=u+atv=u+at

{\boxed {\boxed {v²=u²+2as}}}

v²=u²+2as

s=ut+\frac{1}{2} at²s=ut+

2

1

at²

now\:substitute\:the\:valuesnowsubstitutethevalues

8²=0²+2(2)(s)8²=0²+2(2)(s)

64=0+4\times s64=0+4×s

64=4\times s64=4×s

s=\frac{64}{4}s=

4

64

{\boxed {\boxed {s=m}}}

s=m

\therefore the\:distance \:travelled \:by\:the \:PARTICLES \:is\:{\boxed {\boxed {16m}}}∴thedistancetravelledbytheparticlesis

16m

\blue{\boxed{\blue{ \BOLD{\fcolorbox{red}{black}{\green{☺︎︎Hope\:It\:Helps☺︎︎}}}}}}

☺︎︎HopeItHelps☺︎︎

{\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5069}}}}}}}}}}}}}}}

@suraj5069

\star\underline{\mathtt\orange{⫷❥Q᭄} \mathfrak\blue{u~ }\mathfrak\blue{Σ} \mathbb\purple{ §}\mathtt\orange{T} \mathbb?\pink{iOn⫸}}\star\:⋆

⫷❥Q᭄u Σ§T?iOn⫸

⋆

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?

{ { \underbrace{ \mathbb{ \red{GiVeN\ }}}}}

GiVeN

Initial \:velocity(u) =0m/sInitialvelocity(u)=0m/s

Final\:velocity (v) =8m/sFinalvelocity(v)=8m/s

Time\: taken (t) =4sTimetaken(t)=4s

{ { \underbrace{ \mathbb{ \red{To\:PrOvE\ }}}}}

ToPrOvE

Distance \:travelled \:of \:a \:particleDistancetravelledofaparticle

\star\underbrace{\mathtt\red{⫷❥ᴀ᭄} \mathtt\green{n~ }\mathtt\blue{ §} \mathtt\purple{₩}\mathtt\orange{Σ} \mathtt\pink{R⫸}}\star\:⋆

⫷❥ᴀ᭄n §₩ΣR⫸

⋆

{\boxed {\boxed {v=u+at}}}

v=u+at

v²=u²+2asv²=u²+2as

s=ut+\frac{1}{2} at²s=ut+

2

1

at²

v=Final\:velocityv=Finalvelocity

u=Initial\:velocityu=Initialvelocity

a=accelerationa=acceleration

t=Time\:takent=Timetaken

now\:substitute\:the \:valuesnowsubstitutethevalues

8=0+a(4)8=0+a(4)

8=a\times 48=a×4

a= \frac{8}{4}a=

4

8

{\boxed {\boxed {a=2m/s²}}}

a=2m/s²

________________________________

v=u+atv=u+at

{\boxed {\boxed {v²=u²+2as}}}

v²=u²+2as

s=ut+\frac{1}{2} at²s=ut+

2

1

at²

now\:substitute\:the\:valuesnowsubstitutethevalues

8²=0²+2(2)(s)8²=0²+2(2)(s)

64=0+4\times s64=0+4×s

64=4\times s64=4×s

s=\frac{64}{4}s=

4

64

{\boxed {\boxed {s=m}}}

s=m

\therefore the\:distance \:travelled \:by\:the \:particles \:is\:{\boxed {\boxed {16m}}}∴thedistancetravelledbytheparticlesis

16m

\blue{\boxed{\blue{ \bold{\fcolorbox{red}{black}{\green{☺︎︎Hope\:It\:Helps☺︎︎}}}}}}

☺︎︎HopeItHelps☺︎︎

{\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5069}}}}}}}}}}}}}}}

@suraj5069