Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. How do you determine speed of the vehicle?

Jun 27, 2016

Speed $= \textcolor{red}{40 \text{ miles/hour}}$

Explanation:

Let $s$ be the speed (in miles/hour) that Steve was traveling for $h$ hours to cover $200$ miles.

We are told that if he had traveled at a speed of $\left(s + 10\right)$ miles/hour that it would have taken him $\left(h - 1\right)$ hours to cover the $200$ miles.

Since distance travelled $=$speed $\times$ time

$\textcolor{w h i t e}{\text{XXX}} 200 = s h$
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} \rightarrow \textcolor{b l u e}{h} = \textcolor{g r e e n}{\frac{200}{s}}$
and
color(white)("XXX")200=(s+10)(color(blue)(h)-1))

So we have
$\textcolor{w h i t e}{\text{XXX}} 200 = \left(s + 10\right) \left(\frac{200}{s} - 1\right)$

$\textcolor{w h i t e}{\text{XXXXXX}} = s \left(\frac{200}{s} - 1\right) + 10 \left(\frac{200}{s} - 1\right)$

$\textcolor{w h i t e}{\text{XXXXXX}} = 200 - s + \frac{2000}{s} - 10$

$\Rightarrow \textcolor{w h i t e}{\text{XXX}} s - \frac{2000}{s} + 10 = 0$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} {s}^{2} + 10 s - 2000 = 0$

Using the quadratic formula
$\textcolor{w h i t e}{\text{XXX}} s = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(1\right) \left(- 2000\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 10 \pm \sqrt{8100}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 10 \pm 90}{2}$

$s = - 25$ or $s = 40$

Since the speed must be non-negative, $s = - 25$ is an extraneous solution.

Jun 27, 2016

The slower speed is 40 mph, the faster speed is 50 mph.

Explanation:

There are 2 different scenarios described here, Write an expression for the speed of each of them.

The difference between the times would be 1 hour. This allows us to make an equation.

Let the slower speed be $x$ miles per hour.
The faster speed is $x + 10$ miles per hour.

$t i m e = \text{distance"/"speed}$

At the slower speed, the time, ${T}_{1} = \frac{200}{x}$ hours

At the faster speed, the time, ${T}_{2} = \frac{200}{x + 10}$hours

(${T}_{1} \text{will be more than } {T}_{2}$ because if we drive at slower speed, the journey will take longer.)
The difference between the two times is 1 hour.

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

$\frac{200}{x} - \frac{200}{x + 10} = 1 \text{ now solve the equation}$

Multiply each term by $\textcolor{red}{x \left(x + 10\right)}$

$\frac{\textcolor{red}{x \left(x + 10\right)} \times 200}{x} - \frac{\textcolor{red}{x \left(x + 10\right)} \times 200}{x + 10} = \textcolor{red}{x \left(x + 10\right)} \times 1$

$\frac{\cancel{x} \left(x + 10\right) \times 200}{\cancel{x}} - \frac{x \cancel{\left(x + 10\right)} \times 200}{\cancel{\left(x + 10\right)}} = x \left(x + 10\right) \times 1$

$200 x + 2000 - 200 x = {x}^{2} + 10 x$

${x}^{2} + 10 x - 2000 = 0$

Find factors of 2000 which differ by 10.

The factors must be quite close to $\sqrt{2000}$, because there is a very small difference between them.

We find $40 \times 50 = 2000$

$\left(x - 40\right) \left(x + 50\right) = 0$

$x = 40 \mathmr{and} x = - 50 ,$ (reject -50)

The slower speed is 40mph, the faster speed is 50 mph.