Chapter 5.5, Problem 35E

To determine

To calculate: The time at which the person and his friend met.

Answer to Problem 35E

The time at which the person and his friend met is $18\text{ seconds}$ .

Explanation of Solution

Given information:

The person’s distance d from the fence after t seconds is $d=\left|-5t+100\right|$ and friends distance is $d=\left|-3\frac{1}{3}t+50\right|$ .

Calculation:

Consider the person’s distance d from the fence after t seconds is $d=\left|-5t+100\right|$ and friends distance is $d=\left|-3\frac{1}{3}t+50\right|$ .

They both will meet when distances of both are equated together, so,

  $\left|-5t+100\right|=\left|-3\frac{1}{3}t+50\right|$

Simplify the equation obtained above, apply the property of modulus.

Either $-5t+100=-3\frac{1}{3}t+50$ or $-5t+100=-\left(-3\frac{1}{3}t+50\right)$

Rewrite the above equations in terms of different equations and then use the graph to plot them and verify the solution.

Consider the system of linear equations as,

  $\begin{array}{c}d=-5t+100\\ d=-3\frac{1}{3}t+50\end{array}$

Plot the above system in the coordinate plane. Here $\left(t,d\right)$ is same as $\left(x,y\right)$ .

  

The red curve denotes the line $d=-5t+100$ and blue curve denotes the line $d=-3\frac{1}{3}t+50$ .The graphs intersect at the point $\left(30,-50\right)$ .

Consider the system of linear equations as,

  $\begin{array}{c}d=-5t+100\\ d=3\frac{1}{3}t-50\end{array}$

Plot the above system in the coordinate plane. Here $\left(t,d\right)$ is same as $\left(x,y\right)$ .

  

The red curve denotes the line $d=-5t+100$ and blue curve denotes the line $d=3\frac{1}{3}t-50$ .The graphs intersect at the point $\left(18,10\right)$ .

Since, distance is negative in first case that is $-50\text{ meters}$ so, this case is rejected.

Therefore, the two friends meet after 18 seconds from the starting point.

Thus, the time at which the person and his friend met is $18\text{ seconds}$ .