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. plz help​

Answer 1

Answer:

The distance traveled before takeoff is 1731.3m

Step-by-step explanation:

Given,

the initial velocity = 0 m/s.

acceleration = 3.20 m/s^2

time = 32.8 s

According to laws of motion.

sut + 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 Ans.

I hope it's help with...☺

Answer 2

Answer:

• $\sf{1721.344 m}$

Step-by-step explanation:

$\:\:{\underline{\sf{\:\:\:\:Given\: :-\:\:\:}}}$

• $\sf a =3.20m/s^2$
• $\sf t = 32.8 \:s$
• $\sf v_{i} = 0m/s$

$\:{\underline{\sf{\:\:\:\: Tofind\: :-\:\:\:}}}$

• $\textsf{Distance traveled before takeoff}$

$\:\:\:{\underline{\sf{\:\:\:\: Calculation\: :-\:\:\:}}}$

$\:\:\:\dashrightarrow\:\:\sf{s = ut + \dfrac{1}{2} at^2 }$

$\:\:\:\dashrightarrow\:\:\sf{s = \dfrac{1}{2} at^2}$

$\:\:\:\dashrightarrow\:\:\sf{s= \dfrac{1}{2}(3.20)\times (32.8)^2}$

$\:\:\:\dashrightarrow\:\:\sf{s= 1721.344 m}$

$\bigstar\:{\underline{\textbf{\textsf{\:\:\:Distance\: traveled\: before\: takeoff\: is\: 1721.344 \:m\:\:\:}}}}$