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

Given,
the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m

Answer 2

Answer: The distance traveled before takeoff is 1720 m.

Explanation:

According to the question,

As we know

An airplane accelerates down a runway at 3.20 m/s² for 32.8 s until is finally lifting off the ground.

The initial speed of an airplane  u=0 m/s

Acceleration of the plane  a=3.2 m/s² and time has taken t=3.28 s

So now by using the formula, distance covered $S = ut+ \frac{1}{2} at^{2}$

⇒   S=0(32.8)+0.5(3.2)(32.8)² = 1721.3 ∼ 1720m

Hence,  the distance travelled before takeoff is 1720 m.

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