Question 152596
You can use the distance formula: {{{d = r*t}}} where: d = distance, r = rate (speed), and t = time of travel.
For the outbound trip:
{{{d[1] = r[1]*t[1]}}}
For the return trip:
{{{d[2] = r[2]*t[2]}}}
In this problem, {{{d[1] = d[2]}}} and {{{t[1]+t[2] = 4}}}, so you can write:
{{{r[1]*t[1] = r[2]*t[2]}}}
{{{r[1] = 40}}} and {{{r[2] = 45}}} These are given, so...
{{{40*t[1] = 45*t[2]}}} Substitute: {{{t[2] = 4-t{1]}}}
{{{40*t[1] = 45*(4-t[1])}}} Simplify and solvefor {{{t[1]}}}
{{{40*t[1] = 180-45*t[1]}}} Add {{{45*t[1]}}} to both sides.
{{{85*t[1] = 180}}} Divide both sides by 85.
{{{t[1] = 2.12}}}hours.
Now you can calculate the one-way distance, {{{d[1]}}}
{{{d[1] = 40*t[1]}}}
{{{d[1] = 40*2.12}}}
{{{d[1] = 84.8}}}miles.
This should be equal to {{{d[2]}}}...let's see if it is:
{{{d[2] = r[2]*t[2]}}} Substitute {{{r[2] = 45}}} and {{{t[2] = 4-2.12}}}={{{1.88}}}
{{{d[2] = 45*1.88}}}
{{{d[2] = 84.6}}}miles.
Michael travelled {{{d[1]+d[2] = 84.8+84.6}}}={{{169.4}}}miles.