two cars are traveling along intersecting roads. one car is 200 meters north of the intersection and moving…

two cars are traveling along intersecting roads. one car is 200 meters north of the intersection and moving towards the intersection at 15 m/s, while the other is 130 meters west of the intersection and moving away from the intersection at 18 m/s. 3 seconds later, the cars are getting select an answer at m/s. (enter your answer rounded to 3 decimal places)

two cars are traveling along intersecting roads. one car is 200 meters north of the intersection and moving towards the intersection at 15 m/s, while the other is 130 meters west of the intersection and moving away from the intersection at 18 m/s. 3 seconds later, the cars are getting select an answer at m/s. (enter your answer rounded to 3 decimal places)

Answer

Explanation:

Step1: Determine the new positions of the cars

Let the first - car's initial position be $y_0 = 200$ m (north of the intersection) and its velocity $v_{y}=- 15$ m/s (negative because it is moving towards the intersection). Let the second - car's initial position be $x_0 = 130$ m (west of the intersection) and its velocity $v_{x}=18$ m/s. After $t = 3$ s, the new position of the first car is $y=y_0+v_{y}t=200-15\times3=200 - 45=155$ m. The new position of the second car is $x=x_0 + v_{x}t=130+18\times3=130 + 54=184$ m.

Step2: Use the distance formula $D=\sqrt{x^{2}+y^{2}}$ and differentiate with respect to time $t$

The distance $D$ between the two cars is given by $D=\sqrt{x^{2}+y^{2}}=(x^{2}+y^{2})^{\frac{1}{2}}$. Differentiating both sides with respect to time $t$ using the chain - rule, we get $\frac{dD}{dt}=\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{\sqrt{x^{2}+y^{2}}}$. We know that $x = 184$ m, $\frac{dx}{dt}=18$ m/s, $y = 155$ m, and $\frac{dy}{dt}=-15$ m/s.

Step3: Substitute the values into the derivative formula

Substitute the values into the formula for $\frac{dD}{dt}$: [ \begin{align*} \frac{dD}{dt}&=\frac{184\times18+155\times(- 15)}{\sqrt{184^{2}+155^{2}}}\ &=\frac{3312-2325}{\sqrt{33856 + 24025}}\ &=\frac{987}{\sqrt{57881}}\ &=\frac{987}{240.585}\ &\approx4.095 \end{align*} ]

Answer:

$4.095$