題目

2. An experiment detects that a particle at

(1,0)

(on the

xy

-plane) is moving towards the north at the speed

3

meters per minute. At the same time another particle at

(-2,1)

is moving towards the east at the speed

4

meters per minute.

(a) $(5\%)$ Let $s(t)$ be the distance (in meters) between the two particles in $t$ minutes. $s(t)=\underline{\quad(3)\quad}$ .

(b) $(10\%)$ Suppose at $t=t_0$ the two particles are closest to each other. Then $t_0=\underline{\quad(4)\quad}$ .

解答

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第一粒子向北移動，第二粒子向東移動，因此 $t$ 分鐘後的位置分別是

(1,3t),\qquad (-2+4t,1).

兩者的位移差為 $(3-4t,3t-1)$ ，所以

\begin{align*} s(t) =&\,\sqrt{(3-4t)^2+(3t-1)^2}\\[4mm] =&\,\sqrt{25t^2-30t+10}. \end{align*}

故 (a) 為 $\boxed{\sqrt{25t^2-30t+10}}$ 。

將根號內配方：

25t^2-30t+10=25\left(t-\frac35\right)^2+1.

此量在 $t=\frac35$ 時最小，因此兩粒子最近的時刻為

\boxed{t_0=\frac35}.