题目

Problem

A car moves along a straight horizontal road with constant acceleration $a\,\mathrm{m\,s^{-2}}$ where $a>0$ .

The car is modelled as a particle.

At time $t=0$ , the car passes point $A$ and is moving with speed $u\,\mathrm{m\,s^{-1}}$ .

In the first three seconds after passing $A$ the car travels $20\,\mathrm{m}$ .

In the fourth second after passing $A$ the car travels $10\,\mathrm{m}$ .

The speed of the car as it passes point $B$ is $20\,\mathrm{m\,s^{-1}}$ .

Find the time taken for the car to travel from $A$ to $B$ .

(8)

题目中文翻译

一辆汽车沿水平直路以恒定加速度 $a\,\mathrm{m\,s^{-2}}$ 运动，其中 $a>0$ 。

将汽车视为质点。

在 $t=0$ 时，汽车经过 $A$ 点，速率为 $u\,\mathrm{m\,s^{-1}}$ 。

经过 $A$ 点后的最初三秒内，汽车行驶 $20\,\mathrm{m}$ 。

经过 $A$ 点后的第四秒内，汽车行驶 $10\,\mathrm{m}$ 。

汽车经过 $B$ 点时的速率为 $20\,\mathrm{m\,s^{-1}}$ 。

求汽车从 $A$ 行驶至 $B$ 所需的时间。

(8)

解答

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“第四秒内”指 $t=3$ 到 $t=4$ 的一秒，而不是前四秒。先用前三秒的位移和第四秒内的位移求出初速度、加速度，再用终速度求总时间。

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Let $s(t)$ be the displacement from $A$ after $t$ seconds. Since the acceleration is constant,

s(t)=ut+\frac12at^2.

The displacement in the first three seconds gives

20=3u+\frac92a.\tag{1}

The displacement during the fourth second is $s(4)-s(3)$ , so

\begin{align*} 10=&\,(4u+8a)-\left(3u+\frac92a\right),\\ 10=&\,u+\frac72a.\tag{2} \end{align*}

Multiplying (2) by $3$ and subtracting (1),

\begin{align*} 30-20=&\,\left(3u+\frac{21}{2}a\right)-\left(3u+\frac92a\right),\\ 10=&\,6a,\\ a=&\,\frac53\ \mathrm{m\,s^{-2}}. \end{align*}

Substitution into (2) gives

u=10-\frac72\cdot\frac53=\frac{25}{6}\ \mathrm{m\,s^{-1}}.

If the time from $A$ to $B$ is $t_B$ seconds, then

\begin{align*} 20=&\,\frac{25}{6}+\frac53t_B,\\ t_B=&\,\frac{20-25/6}{5/3}=\boxed{9.5\ \mathrm{s}}. \end{align*}