Question

hep me to find out ​

Answer 1

Answer :

➥ The distance traveled in 12 sec = 320 m

Given :

➤ Intial velocity (u) = 0

➤ Acceleration (a) = 5 m/s²

➤ Time taken (t) = 8 sec

To Find :

➤ Total distance in 12 sec = ?

Solution :

Distance traveled in 8 sec

From second equation of motion

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

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

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

$\tt{: \implies s = 0 + 1 \times 5 \times 4 \times 8}$

$\tt{: \implies s = 1 \times 160}$

$\tt{: \implies \purple{\underline{ \overline{ \boxed{ \green{ \bf{ \: \: s = 160 \: m \: \: }}}}}}}$

Speed after 8 sec

From first equation of motion

$\tt{: \implies v = u + at}$

$\tt{: \implies v = 0 + 5 \times 8}$

$\tt{: \implies v = 0 + 40}$

$\tt{: \implies \purple{ \underline{ \overline{ \boxed{ \green{ \bf{ \: \: v = 40 \: m/ {s}^{2} \: \: }}}}}}}$

Distance traveled in next 4 sec

$\tt{: \implies d = s \times t}$

$\tt{: \implies d = 40 \times 4}$

$\tt{: \implies \purple{ \underline{ \overline{ \boxed{ \green{ \bf{ \: \: d = 160 \: m \: \: }}}}}}}$

Now ,

Total distance in 12 sec

$\tt{: \implies s + d}$

$\tt{: \implies 160 + 160}$

$\tt{: \implies \green{ \underline{ \overline{ \boxed{ \purple{ \bf{ \: \: 320 \: m \: \: }}}}}}}$

Hence, the distance traveled in 12 sec is 320 m.

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Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

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