Question

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

Answer 1

using formula
$s = ut + \frac{1}{2} a {t}^{2}$
we can find it . s=0+1÷2×3.20×(32.8)×32.8=1721.344

Answer 2

1,721.344 Meters

Given:

Acceleration of the Airplane = + 3.2 m/sec^2

Time taken by the airplane to lift of the ground = 32.8 seconds

Initial velocity of the airplane = 0 m/sec

To Find:

Distance Traveled By The Airplane Before Takeoff.

Calculating:

We can use the equation of motion that allows to calculate distance that is:

s = ut + 1/2 at^2

Substituting all the given values to us in this equation we can get:

= (0 m/s) * (32.8 s) + 1/2 * (3.20 m/s^2)*(32.8 s)^2

= 0 + 1/2 x 3,442.688

= 1,721.344 m

Therefore, the distance traveled by the Airplane before takeoff is 1,721.344 meters.