Question

An airplane is flying at 150 m/s when it begins to slow down for landing. The airplane's acceleration rate is -15m/s2 and it does this for 8s. During this deceleration, how far does the plane travel?

Answer 1

The airplane decelerates from an initial speed of 150 m/s at a rate of -15 m/s² over a period of 8 seconds, and during this time, it covers a distance of 720 meters.

Step-by-step explanation:

The question is asking to determine the distance an airplane travels while decelerating from an initial speed of 150 m/s with an acceleration rate of -15 m/s2 over a period of 8 seconds. To solve this, we can use the formula for distance under constant acceleration: d = vit + (1/2)at2, where d is distance, vi is initial velocity, a is acceleration, and t is time.

Substituting the given values, we have:

• Initial velocity, vi = 150 m/s
• Acceleration, a = -15 m/s2
• Time, t = 8 s

Therefore, the distance traveled is:

d = 150 * 8 + (1/2)(-15)(8)2 = 1200 - (1/2)(15)(64) = 1200 - (7.5)(64) = 1200 - 480 = 720 m.

The airplane travels a distance of 720 meters during deceleration.