two cars are moving towards an intersection. one car is 160 meters north of the intersection and moving…

Explanation:

Step1: Calculate new positions

Let the first - car's initial position $y_0 = 160$ m and speed $v_y=- 18$ m/s, the second - car's initial position $x_0 = 190$ m and speed $v_x=-20$ m/s. After $t = 3$ s, $y=y_0+v_y t=160-18\times3=160 - 54 = 106$ m, $x=x_0+v_x t=190-20\times3=190 - 60 = 130$ m.

Step2: Define the distance formula

The distance $z$ between the two cars is given by the Pythagorean theorem $z=\sqrt{x^{2}+y^{2}}$. Differentiating both sides with respect to time $t$ using the chain - rule, we get $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, so $\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$.

Step3: Calculate $z$

$z=\sqrt{x^{2}+y^{2}}=\sqrt{130^{2}+106^{2}}=\sqrt{16900 + 11236}=\sqrt{28136}\approx167.738$ m.

Step4: Substitute values into the derivative formula

We know that $\frac{dx}{dt}=-20$ m/s, $\frac{dy}{dt}=-18$ m/s, $x = 130$ m, $y = 106$ m, and $z\approx167.738$ m. Then $\frac{dz}{dt}=\frac{130\times(-20)+106\times(-18)}{167.738}=\frac{-2600-1908}{167.738}=\frac{-4508}{167.738}\approx - 26.875$ m/s.

Answer:

$26.875$