निवनिर्वासित या शब्दाचा समूहदर्शक शब्द कोणता
Answers
\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