Question

An airplane initially flying at a speed of 60.0 m/s accelerates at 5.0 m/s2 for 600 meters. What is its velocity after this acceleration?​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Initial velocity, u= 60.0 m/s

• Acceleration, a = 5.0 m/s²

• Displacement, s = 600 m

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Final velocity, v

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{We know that :}}$

$\boxed{\sf v^{2} - u^{2} = 2as}$

$(\bf 3rd \: equation \: of \: motion)$

$\underline{\:\textsf{Now, put the given values :}}$

$\rm \dashrightarrow {v}^{2} = {60}^{2} + 2 \times 5 \times 600 \\ \\ \rm\dashrightarrow {v}^{2} = 3600 + 6000 \\ \\ \rm\dashrightarrow {v}^{2} = 9600 \\ \\ \rm \dashrightarrow v = \sqrt{9600} \\ \\ \rm \dashrightarrow v = 97.98 \: m {s}^{ - 1}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{Final velocity is \textbf{97.98m/s}}}.$

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$\;\;\underline{\textbf{\textsf{ Know More:-}}}$

$(\bf 1st\: equation \: of \: motion)$

1.$\boxed{\sf v = u + at}$

$(\bf 2nd \: equation \: of \: motion)$

2.$\boxed{\sf s = ut + \frac{1}{2} at^{2}}$

$(\bf 3rd \: equation \: of \: motion)$

3.$\boxed{\sf v^{2} - u^{2} = 2as}$

Where,

›› s = Distance Covered

›› u = Initial Velocity

›› t = Time taken

›› v = Final Velocity

›› a = Acceleration

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Answer 2

Answer:

acceleration........