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.

Question

anonymous · Answer

Which formula should I use?

goformit100 · Answer

use Vector Formula.

anonymous · Answer

that only tells me A + B -> AB (Resultant)..

anonymous · Answer

I know it would have to be accelerating "due north" at a certain (m/s2).

anonymous · Answer

How do I solve for acceleration please?

anonymous · Answer

$$v=v_0-at$$ 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 · Answer

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 · Answer

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