A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
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50
Question :
A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
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Answer :
vi = 0 m/s
vf = 88.3 m/s
d = 1365 m
Answer to be found :
vf² = vi² + 2*a*d
(88.3 m/s)² = (0 m/s)² + 2*(a)*(1365 m)
7797 m²/s² = (0 m²/s²) + (2730 m)*a
7797 m²/s² = (2730 m)*a
(7797 m²/s²)/(2730 m) = a
a = 2.86 m/s²
vf = vi + axt
88.3 m/s = 0 m/s + (2.86 m/s²)*t
(88.3 m/s)/(2.86 m/s²) = t
t = 30. 8 s
A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
___________________________________________
Answer :
vi = 0 m/s
vf = 88.3 m/s
d = 1365 m
Answer to be found :
vf² = vi² + 2*a*d
(88.3 m/s)² = (0 m/s)² + 2*(a)*(1365 m)
7797 m²/s² = (0 m²/s²) + (2730 m)*a
7797 m²/s² = (2730 m)*a
(7797 m²/s²)/(2730 m) = a
a = 2.86 m/s²
vf = vi + axt
88.3 m/s = 0 m/s + (2.86 m/s²)*t
(88.3 m/s)/(2.86 m/s²) = t
t = 30. 8 s
Answered by
30
Distance = 1365m
Initial velocity = 0(starts from rest)
Final velocity = 88.3m/s
V sqr = u sqr +2as
88.3 sqr = 0 + 2 (1365)a
7796.8 = 2739.a
a = 7796.8/2739
a=2.856m/sec sqr
V= u+at
88.3 = 0+2.856.t
t=2.856/88.3
t=30.917 seconds
Hope u understand
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