An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

To solve this problem, we can use the formula for distance in uniformly accelerated motion:

d = vit + 0.5at^2

where:

• d is the distance,
• vi is the initial velocity,
• a is the acceleration, and
• t is the time.

In this case, the airplane starts from rest, so the initial velocity (vi) is 0. The acceleration (a) is 3.20 m/s^2 and the time (t) is 32.8 s. Substituting these values into the formula gives:

d = 0*(32.8 s) + 0.5*(3.20 m/s^2)*(32.8 s)^2

Solving this equation gives the distance traveled before takeoff.

To solve this problem, we can use the formula for distance in uniformly accelerated motion:

d = vit + 0.5at^2

where:

• d is the distance,
• vi is the initial velocity,
• a is the acceleration, and
• t is the time.

Given that the airplane starts from rest, the initial velocity (vi) is 0. Therefore, the formula simplifies to:

d = 0.5at^2

Substituting the given values:

d = 0.5 * 3.20 m/s^2 * (32.8 s)^2

Now, calculate the value:

d = 0.5 * 3.20 m/s^2 * 1075.84 s^2

d = 1721.344 m

So, the airplane traveled 1721.344 meters before takeoff.