Question

A body starts from rest and moving with constant acceleration. If it covers 20m in first 5 seconds then the distance covered by it in the next 5 seconds is                                                                     (a) 20m                      (b) 40m                                      (c) 60m                                  (d)80m

fastest answer will be marked the brainliest​

Answer 1

Answer :

➥ The distance covered in next 5 sec = 20 m

Given :

➤ Distance covered in first 5 sec = 20 m

➤ Time taken by the body (t) = 5 sec

To Find :

➤ The distance covered in next 5 sec = ?

Solution :

Distance covered 20 m in 5 sec

From second equation of motion

$\tt{: \implies s = ut + \dfrac{1}{2} a {t}^{2} }$

$\tt{: \implies 20 = 0 \times 5 + \dfrac{1}{2} \times a \times 5 \times 5 }$

$\tt{: \implies 20 = 0 + \dfrac{1}{2} \times a \times 25}$

$\tt{: \implies 20 = 0 + \dfrac{1}{2} \times 25a}$

$\tt{: \implies 20 = \dfrac{25a}{2} }$

$\tt{: \implies 40 = 25a}$

$\tt{: \implies \cancel{\dfrac{40}{25}} = a}$

$\tt{: \implies \purple{ \underline{ \overline{ \boxed{ \green{ \bf{ \: \: a = \dfrac{8}{5} \: m/s^2 \: \: }}}}}}}$

Now ,

Distance covered in next 5 seconds.

From second equation of motion

$\tt{: \implies s = ut + \dfrac{1}{2}a {t}^{2}}$

$\tt{: \implies s = 0 \times 5 + \dfrac{1}{ \cancel{2}} \times \dfrac{ \cancel{8}}{ \cancel{5}} \times \cancel{5} \times 5}$

$\tt{: \implies s = 0 + 1 \times 4 \times 5}$

$\tt{: \implies s = 0 + 1 \times 20}$

$\tt{: \implies s = 0 + 20}$

$\tt{: \implies \green{ \underline{ \overline{ \boxed{ \purple{ \bf{ \: \: s = 20 \: m \: \: }}}}}}}$

Hence, the distance covered in next 5 second is 20 m.

⠀

▭▬▭▬▭▬▭▬▭▬▭▬▭▬▭▬▭▬

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

▭▬▭▬▭▬▭▬▭▬▭▬▭▬▭▬▭▬