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.
Answers
Answered by
557
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
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
Answered by
1
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=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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