anonymous
  • anonymous
Landing with a speed of 81.9 m/s, and traveling due south, a jet comes to rest in 949m. Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Which formula should I use?
goformit100
  • goformit100
use Vector Formula.
anonymous
  • anonymous
that only tells me A + B -> AB (Resultant)..

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anonymous
  • anonymous
I know it would have to be accelerating "due north" at a certain (m/s2).
anonymous
  • anonymous
How do I solve for acceleration please?
anonymous
  • anonymous
\[v=v_0-at\] when it comes to rest:\[v=0 \rightarrow t=\frac{ v_0 }{ a }\]distance is calculated like:\[S=v_0t-\frac{ 1 }{ 2}at^2=v_0\left( \frac{ v_0 }{ a} \right)-\frac{ 1 }{ 2 }a\left( \frac{ v_0 }{ a} \right)^2=\frac{ v_0^2 }{ 2a }\rightarrow a=\frac{ v_0^2 }{ 2S }\] You know Vo=81.9 m/sec and S=949
anonymous
  • anonymous
Thank you, I used \[\frac{ v _{0}^{2} }{ 2S }\] v initial = 81.9 m/sec and S=949m to get 3.53 m/sec^2 Can you please go over the derivation again to eliminate t? I do not fully understand the last step.
anonymous
  • anonymous
Just replace the expresion of time to come to rest t=Vo/a in the expression of distance run in this time: S=Vo·t-1/2a·t^2

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