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:
π = π£0π‘ + 0.5ππ‘^2
where π is the distance traveled, π£0 is the initial velocity, π is the acceleration, and π‘ is the time.
In this case, the initial velocity π£0 is 0 m/s (since the airplane starts from rest) and the acceleration π is given as 3.20 m/s^2. The time π‘ is given as 32.8 s.
Plugging in these values into the equation, we get:
π = 0(32.8) + 0.5(3.20)(32.8)^2
Simplifying this equation, we get:
π = 0 + 0.5(3.20)(1074.24)
π = 0 + 0.5(3436.35)
π = 0 + 1718.175
Therefore, the distance traveled before takeoff is 1718.175 meters.