# Bill is planning to drive 280 km. If he were to start his trip 1 hour later than he plans, he would need to drive 10 miles/hour faster to make the trip in the same amount of time. What is the average speed at which Bill plans to drive?

Nov 10, 2016

Average speed was $48.15$ miles/hour (approx.)

#### Explanation:

Let $t$ be the amount of time (in hours) required to travel $280$ miles at the speed $s$ (in miles per hour).
$\textcolor{w h i t e}{\text{XXX}} \rightarrow t = \frac{280}{s}$

We told it would take $\left(t - 1\right)$ hours to travel the $280$ miles at a speed of $\left(s + 10\right)$ miles per hour.

Therefore
$\textcolor{w h i t e}{\text{XXX}} \left(t - 1\right) \cdot \left(s + 10\right) = 280$

...and replacing $t$ with $\left(\frac{280}{s}\right)$ from our initial equation:
$\textcolor{w h i t e}{\text{XXX}} \left(\frac{280}{s} - 1\right) \cdot \left(s + 10\right) = 280$

Simplifying
$\textcolor{w h i t e}{\text{XXX}} 280 + \frac{2800}{s} - s - 10 = 280$

$\textcolor{w h i t e}{\text{XXX}} \frac{2800}{s} - s - 10 = 0$

multiplying by $\left(- s\right)$ and re-arranging into standard form:
$\textcolor{w h i t e}{\text{XXX}} {s}^{2} + 10 s - 2800 = 0$

Applying the quadratic formula (but ignoring the theoretically negative result)
$\textcolor{w h i t e}{\text{XXX}} s = \frac{- 10 + \sqrt{{10}^{2} - 4 \cdot 1 \cdot \left(- 2800\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{\text{XXX}} \approx 48.15072906$ (using calculator)