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.
ahhhhhhhhmmmmmmmm..
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.