Answer:
The distance from rest to takeoff is 6116 meters, and the time from rest to takeoff is 138.9 seconds.
Step-by-step explanation:
To calculate the distance and time from rest to takeoff for a jet transport with a constant acceleration of 0.4g, we can use the equations of motion for constant acceleration. The equations for distance and time are:
Distance: s = v0t + 0.5at^2
Time: t = (v - v0)/a
Where s is the distance, v0 is the initial velocity, t is the time, a is the acceleration, and v is the final velocity. In this case, we are given that the initial velocity is 0 (since the aircraft is at rest), the acceleration is 0.4g, and the final velocity is 200 km/h.
To convert the final velocity to meters per second, we need to first convert the velocity from kilometers per hour to meters per second. One kilometer is equal to 1000 meters, and one hour is equal to 3600 seconds, so 200 km/h is equal to 200 * 1000 / 3600 = 55.56 m/s.
We can then plug these values into the equations of motion to calculate the distance and time from rest to takeoff:
Distance: s = 0 * t + 0.5 * 0.4g * t^2
s = 0.2g * t^2
Time: t = (v - v0)/a
t = (55.56 - 0) / 0.4g
t = 138.9 seconds
Therefore, the distance from rest to takeoff is 0.2g * 138.9^2 = 6116 meters, and the time from rest to takeoff is 138.9 seconds.