Two cars are driving towards an intersection from perpendicular directions. $\newline$ The first car's velocity is $10$ meters per second and the second car's velocity is $6$ meters per second. $\newline$ At a certain instant, the first car is $4$ meters from the intersection and the second car is $3$ 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) $-11$ . $6$ $\newline$ (B) $-10$ . $8$ $\newline$ (C) $-5$ $\newline$ (D) $-\sqrt{136}$

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Grade 7

Solve two-step equations: word problems

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Q. Two cars are driving towards an intersection from perpendicular directions.

\newline

The first car's velocity is

10

meters per second and the second car's velocity is

6

meters per second.

\newline

At a certain instant, the first car is

4

meters from the intersection and the second car is

3

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)

-11

6

\newline

(B)

-10

8

\newline

(C)

-5

\newline

(D)

-\sqrt{136}

Formulating Right Triangle: The distance between the cars can be represented by the hypotenuse of a right triangle, with the sides being the distances of the cars from the intersection.
Applying Pythagorean Theorem: Let's call the distance between the cars ＂d＂. Using the Pythagorean theorem, $d^2 = 4^2 + 3^2$ .
Calculating Distance: Calculating $d$ , we get $d^2 = 16 + 9$ which is $d^2 = 25$ .
Finding Rate of Change: Taking the square root of both sides, $d = \sqrt{25}$ , so $d = 5$ meters.
Using Chain Rule: Now, we need to find the rate of change of $d$ with respect to time, which is $\frac{dd}{dt}$ .
Differentiating Distances: Using the chain rule, $\frac{dd}{dt} = \left(\frac{dd}{dx_1}\right)\left(\frac{dx_1}{dt}\right) + \left(\frac{dd}{dx_2}\right)\left(\frac{dx_2}{dt}\right)$ , where $x_1$ and $x_2$ are the distances of the cars from the intersection.
Determining Velocities: Differentiating $d$ with respect to $x_1$ and $x_2$ , we get rac{dd}{dx_1} = rac{4}{5} and rac{dd}{dx_2} = rac{3}{5}.
Calculating Rate of Change: The rates $\frac{dx_1}{dt}$ and $\frac{dx_2}{dt}$ are the velocities of the cars, which are $-10\,\text{m/s}$ and $-6\,\text{m/s}$ , respectively (negative because they are approaching the intersection).
Calculating Rate of Change: The rates $\frac{dx_1}{dt}$ and $\frac{dx_2}{dt}$ are the velocities of the cars, which are $-10 \, \text{m/s}$ and $-6 \, \text{m/s}$ , respectively (negative because they are approaching the intersection).Plugging in the values, $\frac{d^2d}{dt^2} = \left(\frac{4}{5}\right)(-10) + \left(\frac{3}{5}\right)(-6)$ .
Calculating Rate of Change: The rates $\frac{dx_1}{dt}$ and $\frac{dx_2}{dt}$ are the velocities of the cars, which are $-10 \, \text{m/s}$ and $-6 \, \text{m/s}$ , respectively (negative because they are approaching the intersection).Plugging in the values, $\frac{d^2d}{dt} = \left(\frac{4}{5}\right)(-10) + \left(\frac{3}{5}\right)(-6)$ .Calculating $\frac{d^2d}{dt}$ , we get $\frac{d^2d}{dt} = -8 - 3.6$ which is $\frac{d^2d}{dt} = -11.6$ meters per second.