A plane lands 1200 before the end of a runway with an initial veloicty of 450km/h, but it lands too late. It cannot stop in time and goes beyond the runway. It finally
stops after crashing into a field for a distance of 250m. Assume that the plane
decelerates at 5ms^2 on the runway.
a). Find the velocity of the plane when it goes beyond the runway.
b). Find the acceleration of the plane when it is on the field
c). Find the total time taken for the plane to stop after the landing
My working on (a)
v^2 - (450x1000 / 3600) = 2(-5) (1200)
v^2 = -11875
(show -ve sign, is there something wrong ??)
Plane landing
2015-04-29 6:47 am
回答 (2)
2015-04-29 7:25 am
✔ 最佳答案
(a) 450 km/h = 125 m/sUse equation: v^2 = u^2 + 2a.s
with u = 125 m/s, a = -5 m/s^2, s = 1200 m, v =?
hence, v^2 = 125^2 + 2.(-5).(1200)
v^2 = 3625 (m/s)^2
v = 60.21 m/s (= 217 km/h)
(b) Use: v^2 = u^2 + 2a.s
with v = 0 m/s, u = 60.21 m/s, s = 250 m, a =?
hence, 0 = 60.21^2 + 2a(250)
a = -7.25 m/s^2
(c) Use equation: v = u + at
On the runway: 60.21 = 125 + (-5)t
i.e. t = 12.96 s
On the field: 0 = 60.21 + (-7.25)t
t = 8.305 s
Therefore, total time lapsed on stopping the plane
= (12.96 + 8.305) s
= 21.26 s
2015-04-29 5:20 pm
上面計了答案給你，但沒有回答你的錯在那裏：
(450X1000/3600)你沒有"平方"^2！
(450X1000/3600)你沒有"平方"^2！
收錄日期: 2021-04-11 21:01:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150428000051KK00110