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Space Time & Motion A.1 Kinematics (id: 098ce5419)

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admin 发表于 2024-2-7 02:32:04 | 显示全部楼层 |阅读模式
本题目来源于试卷: Space Time & Motion A.1 Kinematics,类别为 IB物理学

[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:
$a = \frac{v - u}{t} \Rightarrow t = \frac{v - u}{a} = \frac{75.0 \text{ m s}^{-1} - 0}{1.25 \text{ m s}^{-2}} = \frac{75.0 \text{ m s}^{-1}}{1.25 \text{ m s}^{-2}} = 60 \text{ s}$

$v^2 = u^2 + 2as \Rightarrow s = \frac{v^2 - u^2}{2a} = \frac{(75.0 \text{ m s}^{-1})^2 - 0^2}{2(1.25 \text{ m s}^{-2})} = \frac{(75.0 \text{ m s}^{-1})^2}{2(1.25 \text{ m s}^{-2})} = 2250 \text{ m}$

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