A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest. (a from the instant

Question

A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest.
(a from the instant the plane touches the runaway, what is the minimum time needed before it can come to a rest?
(b Can this plane land on a runaway that is only 0.800 km long?
shown work pls will reward alot of points

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Thu Thủy 4 years 2021-07-16T02:28:28+00:00 1 Answers 85 views 0

Answers ( )

  1. kimchi2
    0
    2021-07-16T02:29:33+00:00
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    Answer:

    a)   t = 20 s,  b)  x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance

    Explanation:

    This is a kinematics exercise

    a) in minimum time to stop,

               v = vo + at

               v = 0

               t = -v0 / a

    we calculate

              t = -100 / (5.00)

              t = 20 s

    b) Let’s find the length you need to stop

              v² = vo² + 2 a x

              x = -v0 ^ 2 / 2a

              x = – 100² / 2 (-5.00)

              x = 1000 m

    As the runway is only 800 m long, the plane cannot land at this distance.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )