Question

I starts walking from my home with an acc of 5 m/s^2 and reached my school after 50 s .Calculate the distance of of my school ,

Answer 1

$\large \dag$ Question :-

I starts walking from my home with an acceleration of 5 m/s² and reached my school after 50 s. Calculate the distance of of my school .

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Distance   \: of \: School \: is \: 6.25\: km }} }$

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

❒ We Have 2nd Equation of Motion as :

Here We Have :

• Initial Velocity = u = 0 m/s

• Time Taken = t = 50 s

• Acceleration a = 5 m/s²

• Let Distance be = s m

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

⏩ Applying 2nd Equation of Motion ;

$:\longmapsto \rm s = 0 \times 50 + \frac{1}{2} \times 5 \times {50}^{2}$

$:\longmapsto \rm s = 0 + \frac{1}{2} \times 5 \times 2500$

$:\longmapsto \rm s = 1250 \times 5$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s = 6250\:m} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{\text Distance = 6.25 \: \text km }}}}}$

Answer 2

QUESTION-:

I starts walking from my home with an acc of 5 m/s² and reached my school after 50 s .Calculate the distance of  my school ,

EXPLANATION-:

• initial velocity(u)=0  m/s   (Since the body starts from rest)
• Acceleration(a)= 5 m/s²
• Time taken(t)=50 s
• Distance(S)= ??

Using the 2nd equation of motion-:

$\bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \yellow{ \bf\; s = ut + \frac{1}{2}at}}}}$

Putting values-:

→s=0×50+1/2×25×(50)²

→s=1/2×5×2500

→s=2.5×2500

→s=6250 m

So the distance is 6250 m or 6.25 km

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