1 Answer

Jack Hughes · Answer 1 · 2023-07-14T02:39:19+0000

Answer:

Usual speed:
$500\; {\rm km \cdot h^(-1)}$ .

Step-by-step explanation:

Let the usual duration of the trip be
hours (
t > 0 .) The actual duration of this trip would be
(t - (1/2)) hours.

The usual speed would be:

$\begin{aligned}(1500)/(t)\end{aligned}$ .

The actual speed would be:

$\begin{aligned}(1500)/(t - (1/2))\end{aligned}$ .

The actual speed is
$100\; {\rm km \cdot h^(-1)}$ greater than the usual speed. In other words:

$\begin{aligned}(1500)/(t - (1/2)) - (1500)/(t) = 100\end{aligned}$ .

Simplify this equation and solve for
(
t > 0 ):

$\begin{aligned}(15)/(t - (1/2)) - (15)/(t) = 1\end{aligned}$ .

$\begin{aligned}15\, t - 15\, \left(t - (1)/(2)\right) = t^(2) - (1)/(2)\, t\end{aligned}$ .

$\begin{aligned}t^(2) - (1)/(2)\, t - (15)/(2) = 0\end{aligned}$ .

$2\, t^(2) - t - 15 = 0$ .

$(2\, t + 5)\, (t - 3) = 0$ .

Since
t > 0 , the only possible value for
would be
t = 3 .

Thus, the usual duration of this trip would be
$3\; {\rm h}$ . The usual speed of this trip would be:

$\begin{aligned}v &= (s)/(t) = \frac{1500\; {\rm km}}{3\; {\rm h}} = 500\; {\rm km \cdot h^(-1)}\end{aligned}$ .

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