A jumbo jet starting from rest needs to attain a speed of 350 km/h on the runway to take off. Can this plane use a 1-km long runway if it is to accelerate at the rate of 2.67 m/s2 (meter per second squared)? Please explain why. Thanks.
2007-12-11 18:10:07 · 4 answers · asked by bea 2 in Science & Mathematics ➔ Physics
350 km/h = 97.22 m/s
By equation v^2 - u^2 = 2 a s,
(97.22)^2 - (0)^2 = 2 (2.67) s
9452.16 = 5.34 s
s = 1770 > 1000
1-km long runway is not long enough.
The runway should be at least 1.77 km long.
2007-12-11 18:47:24 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Velocity v = sqrt(2aS); where a = 2.67 m/sec^2 and S = 1,000 m. If V = 350 km/hr = 350,000 m/hr * 1 hr/3600 sec <= v, then the aircraft will take off. You can do the math.
Just eyeballing the numbers V ~ 100 mps > 70 mps = v; so the answer, based on these rough numbers is "No, the A/C will be too slow (by about 30 mps) to take off."
2007-12-11 18:47:53 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
a = 2.67 m/s^2
v = 350 km/h = 97.22 m/s
t = 36.4 s
d = 1/2 at^2 = 1770 m.
We're 770 m too short.
Answer : NO
2007-12-11 18:52:32 · answer #3 · answered by eSsLaEe 1 · 0⤊ 0⤋
350 km = 218 mph
32.17 *.27 = 8.58 ft/s^2
3300ft
-----------------------------------
3300 = 1/2 * 8.58 * t^2
3300 / 1/2 * 8.58 = t^2
(3300 / 1/2 * 8.58)^0.5 = t
t =35 secs
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v= a * t
235
NO
2007-12-11 18:44:48 · answer #4 · answered by JavaScript_Junkie 6 · 0⤊ 1⤋