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The distance given is a "reference" distance. In other words, I don’t know if the plane travels exactly that distance in stopping. To answer this question, you need to solve for deltax and compare it to this reference distance. Since you can’t answer yes or no in utexas, input your solved for deltax in kilometers .To convert meters to km, just divide your answer by 1000. A jet plane lands with a speed of 110 m/s and can decelerate uniformly at a maximum rate of 4.9 m/s² as it comes to rest. Can this plane land at an airport where the runway is 0.83 km long? Answer this by calculating.

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Final answer:

The jet plane requires a stopping distance of 1.23469 km to come to a complete stop, which is greater than the runway length of 0.83 km, so it cannot safely land on this runway.

Step-by-step explanation:

To determine if a jet plane can land on an 0.83 km long runway, we must calculate the stopping distance, or δx, given the initial velocity of 110 m/s and a maximum deceleration rate of 4.9 m/s². Using the kinematic equation v^2 = u^2 + 2as (where v = final velocity, u = initial velocity, a = acceleration, and s = stopping distance), and rearranging it to solve for s, we get s = (v^2 - u^2) / (2a). Since the plane comes to rest, v = 0, thus s = -u^2 / (2a) = - (110 m/s)^2 / (2 * -4.9 m/s²) = -12100 m / -9.8 m/s² = 1234.69 m. Converting this to kilometers, the stopping distance is 1.23469 km. Comparing this to the runway length of 0.83 km, the plane cannot safely land because the stopping distance exceeds the length of the runway.

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