In an emergency an airplane needs to land on a short runway at Albany County Airport. The plane comes in for a landing with a speed of 95 m/s. The planes maximum magnitude of acceleration is 7.07 m/s2 as it comes to a stop. (a) What is the minimum time interval needed for this plane to stop?

t = 13.43 s

Explanation:

In order to find the minimum time required by the plane to stop, we will use the first equation of motion. The first equation of motion is written as follows:

Vf = Vi + at

where,

Vf = Final Velocity of the Plane = 0 m/s (Since, the plane finally stops)

Vi = Initial Velocity of the Plane = 95 m/s

a = deceleration of the plane = - 7.07 m/s²

t = minimum time interval needed to stop the plane = ?

Therefore,

0 m/s = 95 m/s + (- 7.07 m/s²)t

t = (95 m/s)/(7.07 m/s²)

t = 13.43 s

1 m/s^2

Explanation:

V = U + at

20 = 15 + a(5)

5a = 5

acceleration = 1 m/s^2