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At time $t$ hours, where $0 \leq t \leq 3,$ two planes $P$ and $Q$ are cruising at altitudes of

1.1 t^{2} - 2.3 t + 7 miles

and

6 + 2.4 t - 1.2 t^{2} miles

respectively.

Find the total duration, in hours, for which the difference in altitude between the planes is less than $1$ mile, giving your answer to two significant figures.

Hint

To find when $P$ is $1$ mile above $Q$ , you need to solve

(1.1 t^{2} - 2.3 t + 7) - (6 + 2.4 t - 1.2 t^{2}) = 1

You also need to consider the case when $Q$ is above $P$ .

Solution

When $P$ is $1$ mile above $Q$ , we have

\begin{aligned} (1.1 t^{2} - 2.3 t + 7) - (6 + 2.4 t - 1.2 t^{2}) & = 1 \\ 2.3 t^{2} - 4.7 t & = 0 \\ t (2.3 t - 4.7) & = 0 \\ t & \in {0, 2.0434 \dots} \end{aligned}

So $t = 0$ and $t = 2.04$ are two solutions.

When $Q$ is $1$ mile above $P$ , we have

\begin{aligned} (6 + 2.4 t - 1.2 t^{2}) - (1.1 t^{2} - 2.3 t + 7) & = 1 \\ 2.3 t^{2} - 4.7 t - 2 & = 0 \end{aligned}

We need the quadratic formula this time:

\begin{aligned} t & = \frac{4.7 \pm \sqrt{{4.7}^{2} - 4 (2.3) (- 2)}}{(} 4.6) \\ t & \in {0.6041 \dots, 1.4393 \dots} \end{aligned}

$P$ is initially above $Q$ , then below $Q$ , then above $Q$ again. The path of $P$ is convex and the path of $Q$ is concave. The situation must look something like this

Plane paths

The duration when the difference in altitudes is less than $1$ mile is

\begin{array}{r} (0.6041 \dots - 0) + (2.0434 \dots - 1.4393 \dots) = 1.2 hours (3 s.f.) \end{array}