A Boeing 787 is initially moving down the runway at 3.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the engines give the plane a constant acceleration of 1.7 m/s2. The plane then travels a distance of 1900 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne?
d = Vo*t + 0.5a*t^2 = 1900 m.
3t + 0.5*1.7*t^2 = 1900
0.85t^2 + 3t - 1900 = 0
t = 45s.(Use Quadratic Formula).
To find the time it takes for the plane to lift off, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is 3.0 m/s, the acceleration is 1.7 m/s^2, and the distance is 1900 m. We want to find the time it takes for the plane to lift off, so we can rearrange the formula:
distance - (1/2) * acceleration * time^2 = initial velocity * time
1900 - (1/2) * 1.7 * time^2 = 3.0 * time
Now, we can solve for time. Rearranging the equation gives us:
(1/2) * 1.7 * time^2 + 3.0 * time - 1900 = 0
By using the quadratic formula:
time = (-b ± √(b^2 - 4ac)) / (2a)
where a = (1/2) * 1.7, b = 3.0, and c = -1900, we can calculate the time it takes for the plane to lift off.