An airplane accelerates down a runway at 5.20 m/s 2for 32.8 s until it finally lifts off the ground. Determine the distance traveled before takeoff.

Answer 1

To find the distance traveled before takeoff, we can use the equation:

\[d = v_i t + \frac{1}{2}a t^2\]

Where:
\(d\) = distance traveled
\(v_i\) = initial velocity = 0 m/s (since the plane starts from rest)
\(a\) = acceleration = 5.20 m/s^2
\(t\) = time = 32.8 seconds

Plugging in the values, we get:

\[d = 0 + \frac{1}{2}(5.20)(32.8)^2\]
\[d = \frac{1}{2}(5.20)(1075.84)\]
\[d = 2790.464\text{ m}\]

Therefore, the distance traveled before takeoff is 2790.464 meters.