How far does a plane fly in 17 s while its velocity is changing from 144 m/s to 75 m/s at a uniform rate of acceleration?

a = (75-144)/17

s = 144t + 1/2 at^2

To find the distance a plane travels during a certain time while its velocity changes at a uniform rate of acceleration, you need to use the equation of motion:

distance = initial velocity x time + (1/2) x acceleration x time^2

In this case, you have the initial velocity of 144 m/s, the final velocity of 75 m/s, and a time of 17 s. You are also given that the acceleration is uniform, meaning it does not change during this time.

First, let's calculate the acceleration. Since the acceleration is uniform, we can find it by subtracting the initial velocity from the final velocity, and then dividing it by the time taken:

acceleration = (final velocity - initial velocity) / time
= (75 m/s - 144 m/s) / 17 s

Next, calculate the distance traveled using the formula above:

distance = initial velocity x time + (1/2) x acceleration x time^2

Plug in the values and solve:

distance = 144 m/s x 17 s + (1/2) x (75 m/s - 144 m/s) / 17 s x (17 s)^2

Simplifying the equation:

distance = 144 m/s x 17 s + (1/2) x (-69 m/s) / 17 s x (17 s)^2

Once you calculate this expression, you will get the distance traveled by the plane in 17 seconds while its velocity changes from 144 m/s to 75 m/s at a uniform rate of acceleration.