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

Answer:

Correct Question:-

An airplane accelerates down a runaway at $3.20 \: m/s {}^{2}$for 32.8 s until is finally lifts off the groud. Determine the distance travelled before takeoff.

Given:-

v = 0 m/s

a = 3.20 m/s$^{2}$

t = 3.28 s

where 'v' means final velocity;

'a' means acceleration due to gravity

and 't' means time.

To find:-

Distance travelled before takeoff.

Solution:-

According to the third equation of motion,

s = ut + 1/2at$^{2}$

$\Longrightarrow$ s = 0(3.28) + 0.5 × 3.20 × (3.28)$^{2}$

$\Longrightarrow$ s = 0 + 1721.344

$\Longrightarrow$ s = 1721.344 m

Thus, distance travelled before takeoff is 1721.344 m.

Answer 2

Answer:

1721.34 METER IS THE ANS

Explanation: