Two cars are driving away from an intersection in perpendicular directions. $\newline$ The first car's velocity is $5$ meters per second and the second car's velocity is $8$ meters per second. $\newline$ At a certain instant, the first car is $15$ meters from the intersection and the second car is $20$ meters from the intersection. $\newline$ What is the rate of change of the distance between the cars at that instant (in meters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $13$ $\newline$ (B) $9$ . $4$ $\newline$ (C) $25$ $\newline$ (D) $\sqrt{89}$

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Grade 8

Solve equations with variables on both sides: word problems

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Q. Two cars are driving away from an intersection in perpendicular directions.

\newline

The first car's velocity is

5

meters per second and the second car's velocity is

8

meters per second.

\newline

At a certain instant, the first car is

15

meters from the intersection and the second car is

20

meters from the intersection.

\newline

What is the rate of change of the distance between the cars at that instant (in meters per second)?

\newline

Choose

1

answer:

\newline

(A)

13

\newline

(B)

9

4

\newline

(C)

25

\newline

(D)

\sqrt{89}

Calculate Distance: Let's call the distance between the cars ' $d$ '. We can use the Pythagorean theorem to find ' $d$ '. $\newline$ $d^2 = 15^2 + 20^2$ $\newline$ $d^2 = 225 + 400$ $\newline$ $d^2 = 625$ $\newline$ $d = \sqrt{625}$ $\newline$ $d = 25 \text{ meters}.$
Find Rate of Change: Now, we need to find the rate of change of $d$ with respect to time, which is $\frac{dd}{dt}$ . We can use the chain rule where $\frac{dd}{dt} = \frac{dd}{dx_1}\frac{dx_1}{dt} + \frac{dd}{dx_2}\frac{dx_2}{dt}$ , where $x_1$ and $x_2$ are the distances of the first and second car from the intersection, respectively.
Apply Chain Rule: The rates $\frac{dx_1}{dt}$ and $\frac{dx_2}{dt}$ are the velocities of the cars, which are $5\,\text{m/s}$ and $8\,\text{m/s}$ , respectively. $\newline$ So, we need to differentiate ' $d$ ' with respect to ' $x_1$ ' and ' $x_2$ '. $\newline$ Using the Pythagorean theorem, we have $d = \sqrt{x_1^2 + x_2^2}$ . $\newline$ Differentiating both sides with respect to ' $x_1$ ', we get $\frac{dd}{dx_1} = \frac{x_1}{d}$ . $\newline$ Similarly, $\frac{dx_2}{dt}$ $0$ .
Use Chain Rule Equation: Now we plug in the values of $x_1$ , $x_2$ , $\frac{dx_1}{dt}$ , and $\frac{dx_2}{dt}$ into the chain rule equation. $\newline$ $\frac{d}{dt} = \left(\frac{x_1}{d}\right)\left(\frac{dx_1}{dt}\right) + \left(\frac{x_2}{d}\right)\left(\frac{dx_2}{dt}\right)$ $\newline$ $\frac{d}{dt} = \left(\frac{15}{25}\right)(5) + \left(\frac{20}{25}\right)(8)$ $\newline$ $\frac{d}{dt} = \left(\frac{3}{5}\right)(5) + \left(\frac{4}{5}\right)(8)$ $\newline$ $\frac{d}{dt} = 3 + 6.4$ $\newline$ $\frac{d}{dt} = 9.4$ meters per second.