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 timet.

[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?

Short answer

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

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

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

Step-by-step solution

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=2500t2-360t+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.