Physics, 29.08.2019 05:10 vanderrage

6. decelerating a plane at a uniform rate of -8 m/s2, a pilot stops the plane in 484 m. how
fast was the plane going before braking begins?

Answers: 3

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Answer from: Ilcienne9736

88 m/s

Explanation:

If we are talking about an acceleration at a uniform rate, we are dealing with constant acceleration, hence we can use the following equation:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

Where:

$V_{f}=0$ Is the final velocity of the plane (we know it is zero because we are told the pilot stops the plane at a specific distance)

$V_{o}$ Is the initial velocity of the plane

$a=-8m/s^{2}$ is the constant acceleration of the plane

d=484m is the distance at which the plane stops

Isolating $V_{o}$ from (1):

$V_{o}=\sqrt{-2ad}$ (2)

$V_{o}=\sqrt{-2(-8m/s^{2})(484m)}$ (3)

Finally:

$V_{o}=88m/s$ This is the veocity the plane had before braking began

Answer from: Quest

Edit the post so i can see the letters for each answer, and increase the points required so i'd be able to answer it.

Answer from: Quest

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6. decelerating a plane at a uniform rate of -8 m/s2, a pilot stops the plane in 484 m. how
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