Question

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

Answer 1

Answer: 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