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An airplane is rolling down the runway at 15 m/s when it begins to accelerate at 3.5 m/s^2. After 500 meters of this acceleration, how fast is it going?

Sagot :

Jacob193idn

Answer:

Approximately [tex]61\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

In this question, the following information about the motion is given:

  • Acceleration: [tex]a = 3.5\; {\rm m\cdot s^{-2}}[/tex],
  • Displacement during acceleration: [tex]x = 500\; {\rm m}[/tex], and
  • Initial velocity: [tex]u = 15\; {\rm m\cdot s^{-1}}[/tex].

The goal is to find the velocity [tex]v[/tex] after achieving the given displacement of [tex]x = 500\; {\rm m}[/tex]. Since the duration of the acceleration is neither given nor required, the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] would be the most suitable. Rearrange this equation to find [tex]v[/tex] in terms of [tex]u[/tex], [tex]a[/tex], and [tex]x[/tex]:

[tex]\displaystyle v^{2} = u^{2} + 2\, a\, x[/tex].

[tex]\begin{aligned} v &= \sqrt{u^{2} + 2\, a\, x} \\ &= \sqrt{(15\; {\rm m\cdot s^{-1}})^{2} + 2\, (3.5\; {\rm m\cdot s^{-2}})\, (500\; {\rm m})} \\ &\approx 61\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

In other words, the velocity of the aircraft at a displacement of [tex]x = 500\; {\rm m}[/tex] from the initial position would be [tex]v \approx 61\; {\rm m\cdot s^{-1}}[/tex].

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