Question

A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Answer 1

Answer:

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

Explanation:

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

The usual speed would be:

$\begin{aligned}\frac{1500}{t}\end{aligned}$.

The actual speed would be:

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

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

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

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

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

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

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

$2\, t^{2} - t - 15 = 0$.

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

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

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

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