[填空题]
The take off speed op1 z vb-(n wqjwj24t/tf a passenger aircrin/o0 njif,/0u c0 oiyaft is $75.0\,m\,s^{-1}$ . A plane starts from rest and accelerates at a steady rate of $1.25\,m\,s^{-2}$. What is the minimum duration of taxiing $s$, and the minimum length of the runway needed for the take off $m$?
参考答案: 空格1: 60±1
空格2: 2250±5
本题详细解析:
$a = \frac{v - u}{t} \Rightarrow t = \frac{v - u}{a} = \frac{75.0 \text{ m s}^{-1} - 0}{1.25 \text{ m s}^{-2}} = \frac{75.0 \text{ m s}^{-1}}{1.25 \text{ m s}^{-2}} = 60 \text{ s}$
$v^2 = u^2 + 2as \Rightarrow s = \frac{v^2 - u^2}{2a} = \frac{(75.0 \text{ m s}^{-1})^2 - 0^2}{2(1.25 \text{ m s}^{-2})} = \frac{(75.0 \text{ m s}^{-1})^2}{2(1.25 \text{ m s}^{-2})} = 2250 \text{ m}$