# Question a5b62

Nov 6, 2016

$\text{ave speed} = 52.68$mph

#### Explanation:

Let the slower speed be $x$ mph.
The faster speed is therefore $x + 10$ mph

You know the distance and the speeds, write expressions for the each time..
$t = \frac{d}{s}$

${t}_{1} = \frac{280}{x}$ (longer time) $\text{ } \mathmr{and} {t}_{2} = \frac{280}{x + 10}$ (shorter time)

The difference between the two times was $1 \text{ hour}$

${t}_{1} - {t}_{2} = 1$

$\frac{280}{x} - \frac{280}{x + 10} = 1 \text{ } \leftarrow \times L C M$ to cancel the denominators

$\frac{\textcolor{b l u e}{x \left(x + 10\right) \times 280}}{x} - \frac{\textcolor{b l u e}{x \left(x + 10\right) \times 280}}{x + 10} = \textcolor{b l u e}{x \left(x + 10\right) \times} 1$

$280 \left(x + 10\right) - 280 x = x \left(x + 10\right)$

$280 x + 2800 - 280 x = {x}^{2} + 10 x \text{ } \leftarrow$

Now you have a quadratic equation, make it equal to 0.

$0 = {x}^{2} + 10 x - 2800$

This expression does not have factors.

Completing the square gives:

${x}^{2} + 10 x + 25 = 2800 + 25$

${\left(x + 5\right)}^{2} = 2825$

$x + 5 = \sqrt{2825} \text{ }$ only the positive root is valid

$x = \sqrt{2825} - 5$

$x = 48.1507$ mph - this is the slower speed

$x + 10 = 58.1507$ mph -this is the faster speed

${t}_{1} = \frac{280}{48.1507} = 5.815 h r s$

${t}_{2} = \frac{280}{58.1507} = 4.815 h r s$

"ave speed" = ("total distance")/("total time")#

$\text{ave speed} = \frac{560}{10.63}$

$\text{ave speed} = 52.68$mph

An expression for the ave speed would be

$\text{ave speed} = \frac{560}{{t}_{1} + {t}_{2}}$