Question 122253
<pre><big><b>

The basic formula is {{{d=rt}}}.


Let's call the distance travelled by the motorcycle {{{d[1]}}}, the
motorcycle's rate {{{r[1]}}}, and the time spent on the motorcycle {{{t[1]}}}.
Hence, for the motorcycle part of the trip we have {{{d[1]=r[1]t[1]}}}.
Since we are given the motorcycle's speed, we can write {{{d[1]=40t[1]}}}.


Likewise for the car part of the trip we have {{{d[2]=r[2]t[2]}}}, and
substituting the given speed we have {{{d[2]=60t[2]}}}.


We also know that the total distance is 540 miles, so we can say:
{{{d[1]+d[2]=540}}}


And we know that the total time is 11 hours, so we can say:
{{{t[1]+t[2]=11}}}


From here we need to find a way to express a relationship that is an equation
in one variable.


{{{d[1]+d[2]=540}}} => {{{d[2]=540-d[1]}}} and
{{{t[1]+t[2]=11}}}  => {{{t[2]=11-t[1]}}}


Next we can substitute these two expressions for {{{d[2]}}} and {{{t[2]}}}
into the equation that we developed to describe the car part of the trip, thus:


{{{d[2]=60t[2]}}}
{{{540-d[1]=60(11-t[1])}}}


Simplifying and collecting like terms gives us:
{{{540-d[1]=660-60t[1]}}}
{{{-d[1]=660-540-60t[1]}}}
{{{-d[1]=120-60t[1]}}}
{{{d[1]=60t[1]-120}}}


Recalling the equation that we developed for the motorcycle part of the trip:
{{{d[1]=40t[1]}}}, we now have two different expressions for {{{d[1]}}} in terms of {{{t[1]}}} that we can equate, since {{{d[1]=d[1]}}}.


{{{40t[1]=60t[1]-120}}}


Solving:

{{{40t[1]-60t[1]=-120}}}
{{{-20t[1]=-120}}}
{{{t[1]=6}}}

Now we know that the motorcycle part of the trip took 6 of the 11 hours total
time.  Substituting this value into the original equation for the motorcycle
portion we get:

{{{d[1]=40*6=240}}}

Giving us <u>240</u> miles for the distance traveled by motorcycle.

<hr/>
<i>Check the answer</i>

The car part of the trip must have taken {{{11 - 6 = 5}}} hours, so the car
distance must have been {{{60*5=300}}} miles.


{{{240 + 300=540}}} which was the given total distance.  Answer checks.