Chapter 1.9, Problem 15P

To determine

To calculate:

The distance traveled at $80\text{km/h}$

Answer to Problem 15P

The distance can’t be same to negative number. Hence this problem has no solution.

Explanation of Solution

Given information:

The Part of $260$ km trip at $80$ km/h and rest at $100$ km/h

Calculation:

Let, divide the path into the two parts. Assume that the time for the one part as $x$ and the time for another part is $260-x$ .

Use the formula of the time:

  $T=\frac{D}{R}\mathrm{...}\left(1\right)$

Where,

  $T$ : Time

  $D$ : Distance

  $R$ : Rate

For first part:

  $T=\frac{x}{80}$

For second part:

  $T=\frac{260-x}{100}$

The total time for the journey is $2$ hr $30$ min. The equation relating the data:

  $\frac{x}{80}+\frac{260-x}{100}=2\text{hr}30\min$

Then, rewrite $2$ hr $30$ min as $\frac{5}{2}$ :

  $\frac{x}{80}+\frac{260-x}{100}=\frac{5}{2}$

The lower common denominator for equation is $100$ and multiplies the terms by $400$ :

  $\begin{array}{l}\frac{x}{80}\left(400\right)+\frac{260-x}{100}\left(400\right)=\frac{5}{2}\left(400\right)\\ x\left(5\right)+\left(260-x\right)4=5\left(200\right)\end{array}$

Use the distributive property:

  $\begin{array}{l}5x+1040-4x=1000\\ x+1040=1000\end{array}$

Subtract $1040$ from both the sides:

  $\begin{array}{l}x+1040-1040=1000-1040\\ x=-40\end{array}$

The distance can’t be same to negative number. Hence this problem has no solution.