An airplane lands and starts down the runway at a southwest velocity of 41 m/s. What constant acceleration allows it to come to a stop in 1.3 km?

I will be happy to critique your thinking.

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im a stage of asking too.. LOL

31.7m/s^2 northeast

To find the constant acceleration that allows the airplane to come to a stop in 1.3 km, we can use the kinematic equation:

\(v^2 = u^2 + 2as\)

Where:
- v is the final velocity (0 m/s since the airplane comes to a stop)
- u is the initial velocity (41 m/s)
- a is the constant acceleration we want to find
- s is the displacement (1.3 km = 1300 m)

Rearranging the equation, we get:

\(a = \frac{v^2 - u^2}{2s}\)

Substituting the given values into the equation, we have:

\(a = \frac{0^2 - 41^2}{2 \times 1300}\)

Simplifying further:

\(a = \frac{-41^2}{2600}\)

Evaluating the expression:

\(a = \frac{-1681}{2600}\) m/s²

So, the constant acceleration that allows the airplane to come to a stop in 1.3 km is approximately -0.6469 m/s².