# Jiro drives 10km then increases his speed by 10kph and drives another 25km. What is his original speed if the whole ride took 45 minutes (or 3/4 hour)?

Sep 9, 2016

The original speed was $40$ km per hour.

#### Explanation:

With a distance-speed-time problem, remember the relationship:

$s = \frac{d}{t} \text{ }$ Let the original speed be $x$ kph.

We can then write the speeds and times in terms of $x$

$\text{Original speed" = x color(white)(xxxxxxxxxx)"Faster speed} = x + 10$

$\text{distance = "10kmcolor(white)(xxxxxxxxxx)" distance =} 25 k m$

$\rightarrow t i m {e}_{1} = \frac{10}{x} \text{hours} \textcolor{w h i t e}{\times \times \times \times} \rightarrow t i m {e}_{2} = \frac{25}{x + 10}$

The total time for the ride was $\frac{3}{4}$ hour " "(time_1 + time_2)

$\frac{10}{x} + \frac{25}{x + 10} = \frac{3}{4} \text{ } \leftarrow$ now solve the equation

Multiply through by the LCD which is $\textcolor{b l u e}{4 x \left(x + 10\right)}$

$\frac{\textcolor{b l u e}{4 \cancel{x} \left(x + 10\right)} \times 10}{\cancel{x}} + \frac{\textcolor{b l u e}{4 x \cancel{x + 10}} \times 25}{\cancel{x + 10}} = \frac{3 \times \textcolor{b l u e}{\cancel{4} x \left(x + 10\right)}}{\cancel{4}}$

=$40 \left(x + 10\right) + 100 x = 3 x \left(x + 10\right)$

$40 x + 400 + 100 x = 3 {x}^{2} + 30 x \text{ } \leftarrow$ make = 0

$0 = 3 {x}^{2} - 110 x - 400 \text{ } \leftarrow$ find factors

$\left(3 x + 10\right) \left(x - 40\right) = 0$

If $3 x + 10 = 0 \text{ } \rightarrow x = - \frac{10}{3}$ reject negative speed

if$x - 40 = 0 \text{ } \rightarrow x = 40$

The original speed was $40$ km per hour