Q. 19 - Precalculus Enhanced with Graphing Utilities

Question

Two cars are approaching an intersection. One is $2$ miles south of the intersection and is moving at a constant speed of $30$ miles per hour. At the same time, the other car is $3$ miles east of the intersection and is moving at a constant speed of $40$ miles per hour.

(a) Build a model that expresses the distance d between the

cars as a function of time t.

[Hint: At $t=0$, the cars are $2$ miles south and $3$ miles

east of the intersection, respectively.]

(b) Use a graphing utility to graph $d=d\left(t\right)$. For what value of t is d smallest?

Step-by-Step Solution

Answer

Part (a) The distance d between the cars as a function of time t is $d\left(t\right)=\sqrt{2500t^2-360t+13}$.

Part (b) The graph of $d=\sqrt{2500t^2-360t+13}$ is

From the graph, d is smallest when $t=\frac{9}{125}$.

Part (a) Step 1 . Using Pythagorean Theorem in the triangle formed by distance travelled in time t , displacement along x and displacement along y .

Displacement along x$=(2-30t)$

Displacement along y$=(3-40t)$

Using Pythagorean theorem,

$$\begin{gathered} d=\sqrt{{(2-30t)}^2+{(3-40t)}^2} \\ =\sqrt{4+900t^2-120t+9+1600t^2-240t} \\ =\sqrt{2500t^2-360t+13} \end{gathered}$$

Part (b) Step 1. Draw the graph of d = 2500 t 2 - 360 t + 13 .

The graph of $d=\sqrt{2500t^2-360t+13}$ is

From the graph, d is smallest for $t=0.07=\frac{9}{125}$hour.