星期一，Dev 跑了 3.6 km。 他在前 1.2 km 的平均速度是 $x$ km/h。 化简并写出一个用 $x$ 表示的在前 1.2 km 所用时间的表达式。 答案 ... 分钟 他在最后 2.4 km 的平均速度比前 1.2 km 慢 2 km/h。 化简并写出一个用 $x$ 表示的在最后 2.4 km 所用时间的表达式。 答案 ... 分钟 Dev 跑完整个 3.6 km 的路程共花了 25 分钟。 列出一个关于 $x$ 的方程，并证明它可以化简为 $25x^{2} - 266x + 144 = 0.$ 解方程 $25x^{2} - 266x + 144 = 0$，将你的答案保留到 3 位小数。 答案 x = ... 星期五，Dev 完成了一次 4 km 的跑步，其平均速度与他在星期一跑 3.6 km 中的最后 2.4 km 时的平均速度相同。 计算 Dev 在星期五跑 4 km 所花的时间。
用分钟和秒给出你的答案，精确到最接近的秒。 答案 ...

1. Find the time for the first 1.2 km and convert it to minutes.

$\text{Time}=\frac{\text{distance}}{\text{speed}}=\frac{1.2}{x}\text{ hours}$
Since $1$ hour $=60$ minutes,
$\frac{1.2}{x}\times 60=\frac{72}{x}\text{ minutes}$

1. Express the time for the final 2.4 km in minutes using the slower speed.

Speed for the final 2.4 km is $(x-2)$ km/h.

$\text{Time}=\frac{2.4}{x-2}\text{ hours}$

Convert to minutes:

$\frac{2.4}{x-2}\times 60=\frac{144}{x-2}\text{ minutes}$

1. Form the equation by adding the two running times to get the total time.

From parts (a)(i) and (a)(ii),

Time for first $1.2$ km $=\dfrac{72}{x}$ minutes,

Time for final $2.4$ km $=\dfrac{144}{x-2}$ minutes.

Since the total time is $25$ minutes,
$\frac{72}{x}+\frac{144}{x-2}=25.$

2. Clear fractions and simplify to the required quadratic equation.

Multiply throughout by $x(x-2)$:
$72(x-2)+144x=25x(x-2).$

Expand and collect terms:
$72x-144+144x=25x^2-50x$
$216x-144=25x^2-50x$
$0=25x^2-266x+144.$

Therefore,
$25x^2-266x+144=0.$

1. Identify the coefficients and apply the quadratic formula.

For $25x^{2}-266x+144=0$,
$a=25,\quad b=-266,\quad c=144.$
Using
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a},$
$x=\frac{-(-266)\pm\sqrt{(-266)^2-4(25)(144)}}{2(25)}.$

2. Evaluate the discriminant and simplify the expression for the roots.

$x=\frac{266\pm\sqrt{70756-14400}}{50}$
$x=\frac{266\pm\sqrt{56356}}{50}.$

3. Calculate both roots and round to 3 decimal places.

$x=\frac{266+\sqrt{56356}}{50}=10.070\ldots$
$x=\frac{266-\sqrt{56356}}{50}=0.569\ldots$
Correct to 3 decimal places,
$x=10.070\text{ or }0.570.$

1. Choose the physically valid value of $x$ and find the Friday running speed.

From part (a)(iv), $x=0.570$ or $x=10.070$.

Since the speed for the last $2.4$ km was $x-2$, $x=0.570$ would give a negative speed, so it is rejected.

Hence
$x=10.070$
Friday speed $=10.070-2=8.070$ km/h.

2. Calculate the time for 4 km and convert it to minutes and seconds.

$\text{Time}=\frac{4}{8.070}\text{ h}$
$=\frac{4}{8.070}\times 60\text{ min}$
$=29.739\ldots\text{ min}$

$0.739\ldots\times 60=44.3\ldots$ seconds.

So the time is approximately $29$ minutes $44$ seconds, to the nearest second.