Chapter 9.8, Problem 11P

To determine

To find: the rate of

Answer to Problem 11P

The person A is walking with the rate $4.713km/h$ and that the person B is walking with the rate $3.713km/h$ .

Explanation of Solution

Given:

Two people walk along perpendicular paths.

One person walks 1 km/h faster than the other.

They are 6 km apart after one hour.

Calculation:

Find the rate at which each person walks.

Suppose that the person A is walking with the rate $xkm/h$ and that the person B is walking with the rate $ykm/h$ .

The person A is walking faster than the person B.

  

Since one person walks 1 km/h faster than the other, therefore,

  $x=y+\mathrm{1........}\left(A\right)$

Since they are 6 km apart after one hour, therefore

  $\sqrt{x^2+y^2}=\mathrm{6.......}\left(B\right)$

Again substituting $x=y+1$ in the equation (B),

  $\begin{array}{l}\sqrt{x^2+y^2}=6\\ x^2+y^2=36\\ {\left(y+1\right)}^2+y^2=36\left[\text{Replacing }x=y+1\right]\\ y^2+2y+1+y^2=36\\ 2y^2+2y-35=0\\ y=\frac{-2\pm \sqrt{4-4\times 2\times \left(-35\right)}}{4}\\ =\frac{-2\pm \sqrt{284}}{4}\\ =\frac{-2\pm 2\sqrt{71}}{4}\\ =\frac{-1\pm \sqrt{71}}{2}\\ =3.713\end{array}$

Since y is rate, therefore only positive value is taken.

Now, substituting y value in the equation (A),

  $\begin{array}{l}x=y+1\\ =3.713+1\\ =4.713\end{array}$

Therefore the person A is walking with the rate $4.713km/h$ and that the person B is walking with the rate $3.713km/h$ .

Conclusion:

The person A is walking with the rate $4.713km/h$ and that the person B is walking with the rate $3.713km/h$ .