Question

two cars are traveling along intersecting roads. one car is 150 meters north of the intersection and moving towards the intersection at 19 m/s, while the other is 140 meters west of the intersection and moving away from the intersection at 11 m/s. 3 seconds later, the cars are getting select an answer at m/s. (enter your answer rounded to 3 decimal places)

Explanation:

Step1: Define Variables and Relationship

Let \( x \) be the distance of the west - moving car from the intersection, \( y \) be the distance of the north - moving car from the intersection, and \( z \) be the distance between the two cars. By the Pythagorean theorem, we have \( z^{2}=x^{2}+y^{2} \).

We know that the initial distance of the north - moving car from the intersection \( y_{0} = 150\) m and it is moving towards the intersection at a speed of \( v_{y}=- 19\) m/s (negative because \( y \) is decreasing). The initial distance of the west - moving car from the intersection \( x_{0}=140\) m and it is moving away from the intersection at a speed of \( v_{x} = 11\) m/s (positive because \( x \) is increasing).

After \( t = 3\) seconds, we can find the values of \( x\), \( y\) and their rates of change.

First, find \( x\) and \( y\) at \( t = 3\):
\( x=x_{0}+v_{x}t=140 + 11\times3=140 + 33=173\) m
\( y=y_{0}+v_{y}t=150-19\times3=150 - 57 = 93\) m

The rate of change of \( x\) with respect to time is \( \frac{dx}{dt}=11\) m/s and the rate of change of \( y\) with respect to time is \( \frac{dy}{dt}=- 19\) m/s.

Step2: Differentiate the Pythagorean Equation

Differentiate \( z^{2}=x^{2}+y^{2}\) with respect to time \( t\):
\( 2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\)
We can simplify this to \( z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

First, find \( z\) at \( t = 3\) using the Pythagorean theorem: \( z=\sqrt{x^{2}+y^{2}}=\sqrt{173^{2}+93^{2}}=\sqrt{29929 + 8649}=\sqrt{38578}\approx196.413\) m

Step3: Solve for \(\frac{dz}{dt}\)

Substitute \( x = 173\), \( y = 93\), \( \frac{dx}{dt}=11\), \( \frac{dy}{dt}=-19\) and \( z\approx196.413\) into the equation \( z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\):

\( 196.413\times\frac{dz}{dt}=173\times11+93\times(- 19)\)

First, calculate the right - hand side:
\( 173\times11 = 1903\)
\( 93\times(-19)=-1767\)
\( 1903-1767 = 136\)

Then, solve for \( \frac{dz}{dt}\):
\( \frac{dz}{dt}=\frac{136}{196.413}\approx0.692\) m/s

Since \( \frac{dz}{dt}>0\), the distance between the two cars is increasing, which means the cars are getting farther apart.

Answer:

farther apart, \( 0.692\)