Question

Suppose a particular jet needs to attain a speed of 220 mph to take off. If it can accelerate from 0 to 220 mph in 45 seconds, how long must the runway be? Assume constant acceleration. Note: 1 mph = 22/15 ft/sec.

Answer 1

The problems states that the acceleration is constant, so y''=a.
Where a is just some constant value.
To find velocity we need to take the anti derivative of both sides with respect to time to get y' = c+at. (Notice that if you solve for 'a' you get (323ft-0)/sec/45sec=a, a=7.17ft/sec^2)
Where y' is your final velocity, c is the initial velocity and at is your acceleration over time.
However you are trying to find distance so you need to take the anti derivative once more to get y=d+c+(1/2)(a)(t^2)
Where y is your final position, d is your initial position, c is the initial velocity and at is your acceleration over time. Since you want the length of the runway, that is the total distance traveled, you subtract your initial position 'd' from your final position.
y-d=c+(1/2)at^2.

Now plug in the values you know:
initial position = 0
final position = y
initial velocity = 0
final velocity = 323
acceleration = 7.17
time = 45 \[y-0=0+\frac{1}{2}*7.17*45^2 \rightarrow y=7259.6\]
Assuming I did my math correctly, your runway needs to be greater than 7260 feet.