An airplane accelerates down a runway at 3.20 m/s
2
for 32.8 s until is finally lifts off the ground. Determine the distancetraveled before takeoff.
X = Vo*t+ 1/2*a*t^2
solve for x
1826.304
1596.3424375m
To determine the distance traveled before takeoff, we can use the equation for distance traveled during constant acceleration:
š = š£0š” + 0.5šš”^2
where š is the distance traveled, š£0 is the initial velocity, š is the acceleration, and š” is the time.
In this case, the initial velocity š£0 is 0 m/s (since the airplane starts from rest) and the acceleration š is given as 3.20 m/s^2. The time š” is given as 32.8 s.
Plugging in these values into the equation, we get:
š = 0(32.8) + 0.5(3.20)(32.8)^2
Simplifying this equation, we get:
š = 0 + 0.5(3.20)(1074.24)
š = 0 + 0.5(3436.35)
š = 0 + 1718.175
Therefore, the distance traveled before takeoff is 1718.175 meters.