# Two bicyclists ride in opposite directions. The speed of the first bicyclist is 5miles per hour faster than the second. After 2hours they are 70miles apart. How do you find their rates?

Sep 19, 2015

${v}_{1} = 20 m p h$
${v}_{2} = 15 m p h$

#### Explanation:

$\Delta v = 5 m p h$
${v}_{1} = {v}_{2} + \Delta v$
${t}_{1} = {t}_{2} = t = 2 h$
$s = 70 m$

v_1=?, v_2=?

${s}_{1} = {v}_{1} \cdot t = \left({v}_{2} + \Delta v\right) \cdot t = {v}_{2} \cdot t + \Delta v \cdot t$
${s}_{2} = {v}_{2} \cdot t$
$s = {s}_{1} + {s}_{2} = {v}_{2} \cdot t + \Delta v \cdot t + {v}_{2} \cdot t$
$s = 2 \cdot {v}_{2} \cdot t + \Delta v \cdot t$
$2 \cdot {v}_{2} \cdot t = s - \Delta v \cdot t$

${v}_{2} = \frac{s - \Delta v \cdot t}{2 \cdot t}$

${v}_{2} = \frac{70 m - 5 m p h \cdot 2 h}{2 \cdot 2 h} = \frac{70 m - 10 m}{4 h} = \frac{60 m}{4 h}$

${v}_{2} = 15 m p h$

${v}_{1} = 15 m p h + 5 m p h = 20 m p h$

Sep 19, 2015

First bicyclist $20$ $m p h$

Second bicyclist $15$ $m p h$

#### Explanation:

Let $x + 5$ $m p h$ be the rate of the first bicyclist

Let $x$ $m p h$ be the rate of the second bicyclist

We know that distance is equal to rate multiplied by the time

$D = r t$

The distance the first bicyclist travels is

${D}_{1} = \left(x + 5\right) \left(2\right)$

The distance the second bicyclist travels is

${D}_{2} = x \left(2\right)$

We are given that they travel $70$ miles apart in the 2 hours

So

${D}_{1} + {D}_{2} = 70$

$\left(x + 5\right) \left(2\right) + x \left(2\right) = 70$

$2 \left(x + 5\right) + 2 x = 70$

$2 x + 10 + 2 x = 70$

$4 x + 10 = 70$

$4 x = 60$

$x = 15$ $m p h$ for the second bicyclist

$x + 5 = 15 + 5 = 20$ $m p h$ for the first bicyclist