Question

Mr. Smith traveled to a city 190 miles from his home to attend a meeting. Due tocar trouble, his average speed returning was 12 mph less than his speed going. Ifthe total time for the round trip was 7 hours, at what rate of speed did he travel tothe city? (Round your answer to the nearest tenth.)

Answer 1

Total distance (d)=190 miles.

Let, r be the rate of speed. and the going speed of the car is r mps.

and time is,

$\begin{gathered} \text{time}=\frac{dis\tan ce}{\text{speed}} \\ t=\frac{190}{r} \end{gathered}$

For returning ,

$\begin{gathered} \text{speed =r-12} \\ \text{time}=\frac{190}{r-12} \end{gathered}$

Total time is 7 hours.

So, it can be written as,

$\begin{gathered} \frac{190}{r}+\frac{190}{r-12}=7 \\ \frac{(r-12)190+190r}{r(r-12)}=7 \\ 190r-2280+190r=7r^2-84r \\ 7r^2-464r+2280=0 \\ \text{compare with ax}^2+bx+c=0 \\ a=7,b=-464,\text{ c=2280} \\ r=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ r=\frac{-\left(-464\right)\pm\sqrt{\left(-464\right)^2-4\cdot\:7\cdot\:2280}}{2\cdot\:7} \\ r=\frac{-\left(-464\right)\pm\:4\sqrt{9466}}{2\cdot\:7} \\ r=\frac{2\left(116+\sqrt{9466}\right)}{7},\: r=\frac{2\left(116-\sqrt{9466}\right)}{7} \\ r=60.94\text{ , r=}5.34 \end{gathered}$

Now, check both the values of r ,For r=60.94

$\begin{gathered} \frac{190}{r}+\frac{190}{r-12}=\frac{190}{60.94}+\frac{190}{60.94-12} \\ =7 \end{gathered}$

So, the speed of car is 60.94 mps.