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A plane suffered an acceleration of 38.5 m/s² that it departed from its base in Cagayan and that it traveled 137.820 km, calculate its final speed during this journey?​

Sagot :

Answer:

According to the data provided, the plane reaches a surprising and unattainable speed of approximately 3257.63 m/s.

Explanation:

The motion being uniformly varied, the acceleration is constant.

Statement data:

[tex]\begin{gathered}\Large\displaystyle\begin{cases} \sf a = 38{,}5 \: m/s^2 \\\sf d = 137{,} 820 \: km \times 1000 = 137820 \: m \\\sf v_0 = 0 \: m/s\\\sf v= \: ? \: m/s \end{cases}\end{gathered}[/tex]

The plane starts from rest and without the time quantity, we use Torricelli's equation.

[tex]\Large\displaystyle\text{${ \sf v^2 = v_{0}^2 +2\cdot a \cdot d}$}[/tex]

[tex]\begin{gathered}\Large\displaystyle\text{${ \sf v^2 = 0^2 +2\cdot 38{,}5 \cdot 137820}$}\\\\\Large\displaystyle\text{${ \sf v^2 = 10612140}$}\\\\\Large\displaystyle\text{${ \sf v = \sqrt{10612140}}$}\\\\\Large\displaystyle\boxed{ \sf v \approx 3257{,}63 \: m/s}\end{gathered}[/tex]

#Hope it helps <3