Math Help for Section 3.5, Page 142
Additional
Example
On a 15-mile trip, a cyclist averages 18 miles per hour before getting
a flat tire. The cyclist walks at an average speed of 3 miles per
hour for the remainder of the trip. The total trip takes 1 hour and 15
minutes. Find the time spent cycling and the time spent walking.
Solution
| Verbal Model: |
$\boxed{\eqalign{&\text{Distance traveled}\cr &\text{cycling}}}\;+\;\boxed{\eqalign{&\text{Distance traveled}\cr &\text{walking}}}\;=\;\boxed{\eqalign{&\text{Total distance}\cr &\text{traveled}}}$ |
|||
| Labels: | Rate (cycling)$\;=18$ | (miles per hour) | ||
| Time spent cycling$\;=t$ | (hours) | |||
| Rate (walking)$\;=3$ | (miles per hour) | |||
| Time spent walking$\;=1.25-t$ | (hours) | |||
| Total distance traveled$\;=15$ | (miles) | |||
| Equation: | $\eqalign{18t+3(1.25-t)=&15}$ | |||
The
formula, $\text{Distance}=\text{Rate}\bullet\text{Time}$, was
used to find the distances in the verbal model. The distance traveled
while cycling is equal to the cycling rate multiplied by the time spent
cycling, and the distance traveled while walking is equal to the walking
rate multiplied by the time spent walking. You can solve the above
equation as follows.
$\eqalign{18t+3(1.25-t)=&15
&{\small\color{red}\quad\quad\text{Write original equation.}}
\cr 18t+3.75-3t=&15
&{\small\color{red}\quad\quad\text{Distributive
Property.}}
\cr 15t+3.75=&15
&{\small\color{red}\quad\quad\text{Combine like terms.}}
\cr 15t=&11.75
&{\small\color{red}\quad\quad\text{Subtract 3.75 from each
side.}}
\cr t=&0.75
&{\small\color{red}\quad\quad\text{Divide each side by 15.}} }$
So, the time spent cycling is $t=0.75$ hours, or 45 minutes, and the
time spent walking is $1.25-t=0.5$ hours, or 30 minutes.
Checking
a Solution
You should check your solution in the original statement of the
problem. Using the calculated times, the total distance traveled
is
$\eqalign{ \left(18\;{\text{miles}\over\cancel{\text{hour}}}\right)(0.75\cancel{\text{
hours}})+\left(3\;{\text{miles}\over\cancel{\text{hour}}}\right)(0.5\cancel{\text{
hours}}) =&13.5\text{ miles}+1.5\text{ miles} \cr
=&15\text{ miles.} {\small\color{red}\quad\text{Solution checks.
}\checkmark}}$