Question

a body starting from the rest moves with a uniform acceleration of 0.02 m/s² for 5 minutes. calculate the speed acquired and the distance travelled in this time.​

Answer 1

$\large \dag$ Question :-

A body starting from the rest moves with a uniform acceleration of 0.02 m/s² for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

$\large \dag$ Answer :-

$\pink{\begin{cases} \bf Speed\:Acquired = 6 \:m/s \\  \\  \bf Distance\:Travelled = 900 \: m\end{cases}}$

$\large \dag$ Step by step Explanation :-

❒ Converting Time in Seconds :-

We Have,

$\text{Time= 5 mins}$

$= \rm( 5 \times 60)second$

$\red{:\longmapsto \text{Time = 300 \:s}}$

Now Here we have :

• Initial Velocity = u = 0 m/s

• Acceleration = a = - 0.02 m/s²

• Time = t = 300 s

1 》 Calculating Speed Acquired :-)

Let Speed Acquired by Body be = v m/s

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{v = u + at}}}$

⏩ Applying 1st Equation of Motion ;

$:\longmapsto \rm v = 0 + 0.02 \times 300$

$:\longmapsto \rm v = 0.02 \times 300$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf v = 6 \: m/s} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{\text Speed \:  \text Acquired = 6\: m/s }}}}}$

2 》Calculating Distance Traveled :-)

❒ We Have 2nd Equation of Motion as :

$\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}}$

⏩ Applying 2nd Equation of Motion :

$:\longmapsto \rm s = 0 \times 300 +  \frac{1}{2}  \times 0.02 \times  {300}^{2}$

$:\longmapsto \rm 0 +  \frac{1}{2}  \times 0.02 \times  {300}^{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s = 900 \: m} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{Distance \:  \text Traveled  = 900 \: m }}}}}$

Answer 2

$\frak{Given}\:\begin{cases}\:\quad\sf Initial \:Velocity \:of\:a\:body\:,\:u\:=\:\pmb{\frak{ 0\:m/s\:}}\:\\\:\quad\sf Acceleration \:of\:a\:body\:,\:a\:=\:\frak{\pmb{ 0.02\:m/s^2\:}}\:\\\:\quad\sf Time\:Taken \:by\:a\:body\:,\:t\:=\:\pmb{\frak{ 5\:min\:}}\:\:\end{cases}\:$

Need To Find : (I) Speed acquired and , (II) Distance Travelled by a body ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

We've the , Initial Velocity (u) , Acceleration (a) and Time Taken by a body , and we'll find Speed Acquired ( or Final Velocity , v ) of the body using First Equation of Motion and that is : $\:\underline {\pmb{\sf v \:=\: u + at\:}}$

$:\implies \sf \:v\:=\:u \:+\: at\:\\\\\\ :\implies \sf \:v\:=\:0 \:+\: (\:0.02\:\times \:5\:min\:)\:\\\\\\ :\implies \sf \:v\:=\:0 \:+\: (\:0.02\:\times \:5\times 60\:)\:\qquad \Bigg\lgroup \:1 \:min \:=\:60\:sec\:\Bigg\rgroup \\\\\\ :\implies \sf \:v\:=\:0 \:+\: (\:0.02\:\times \:5\times 60 \:)\:\\\\\\ :\implies \sf \:v\:=\:0 \:+\: (\:0.02\:\times \:300 \:)\:\\\\\\:\implies \sf \:v\:=\:0.02\:\times \:300 \:\\\\\\ :\implies \pmb {\underline {\boxed {\purple {\:\frak{ \:v\:\:=\:6\:m/s\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf \: Hence,\:Speed\:acquired\:by\:body\:is\:\pmb{\sf 6\:m/s\:}\:.}\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\\\bigstar \:\underline {\sf According \:To\:The\:given\:Question \:\::\:}\:$

Now , We'll find the Distance Travelled (s) using the Third Equation of Motion and that is : $\:\underline {\pmb{\sf s \:=\: ut\:+\:\dfrac{1}{2}\:at^2\:}}$

$\\ \dashrightarrow \sf s \:=\: ut\:+\:\dfrac{1}{2}\:at^2\:\\\\\\ \dashrightarrow \sf s \:=\: (\:0\:\times 5 \:min\:)\:+\:\dfrac{1}{2}\:\times\: 0.02\:(\: 5 \:min\:)^2\:\\\\\\ \dashrightarrow \sf s \:=\: 0\:+\:\dfrac{1}{2}\:\times 0.02\: (\: 5 \:min\:)^2\:\\\\\\ \dashrightarrow \sf s \:=\:\dfrac{1}{2}\:\times 0.02\: (\: 5 \times 60\:)^2\:\:\qquad \Bigg\lgroup \:1 \:min \:=\:60\:sec\:\Bigg\rgroup\\\\\\ \dashrightarrow \sf s \:=\:\dfrac{1}{2}\:\times\: 0.02\: (\:300\:)^2\:\\\\\\ \dashrightarrow \sf s \:=\: 0.01\times 90000\:\\\\\\ \dashrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:s\:\:=\:900\:m\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf \: Hence,\:Distance \:Travelled \:by\:body\:is\:\pmb{\sf 900\:m\:}\:.}\:$