Two cars are driving towards an intersection from perpendicular directions. $\newline$ The first car's velocity is $2$ meters per second and the second car's velocity is $9$ meters per second. $\newline$ At a certain instant, the first car is $8$ meters from the intersection and the second car is $6$ 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) $-8$ . $4$ $\newline$ (B) $-7$ $\newline$ (C) $-10$ $\newline$ (D) $-\sqrt{85}$

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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

2

meters per second and the second car's velocity is

9

meters per second.

\newline

At a certain instant, the first car is

8

meters from the intersection and the second car is

6

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)

-8

4

\newline

(B)

-7

\newline

(C)

-10

\newline

(D)

-\sqrt{85}

Initial Distance Calculation: First car's distance from intersection is $8$ meters, second car's distance is $6$ meters. We need to find the rate at which the distance between them is changing.
Pythagoras' Theorem Application: Let's use Pythagoras' theorem to find the initial distance between the cars. $\text{Distance}^2 = 8^2 + 6^2$ .
Calculation of Initial Distance: Calculating the initial distance: $\text{Distance}^2 = 64 + 36$ .
Differentiation for Rate of Change: $\text{Distance}^2 = 100$ , so $\text{Distance} = \sqrt{100}$ .
Velocities of the Cars: Distance = $10$ meters. This is the initial distance between the cars.
Substitution of Values: Now, let's differentiate the distance with respect to time to find the rate of change. If $x$ is the distance of the first car and $y$ is the distance of the second car from the intersection, then using the chain rule, $\frac{d(\text{Distance})}{dt} = \left(\frac{dx}{dt} * \frac{x}{\text{Distance}}\right) + \left(\frac{dy}{dt} * \frac{y}{\text{Distance}}\right)$ .
Rate of Change Calculation: The velocities of the cars are given as $\frac{dx}{dt} = -2 \, \text{m/s}$ (since it's approaching the intersection, we take it as negative) and $\frac{dy}{dt} = -9 \, \text{m/s}$ (also approaching, hence negative).
Rate of Change Calculation: The velocities of the cars are given as $\frac{dx}{dt} = -2 \, \text{m/s}$ (since it's approaching the intersection, we take it as negative) and $\frac{dy}{dt} = -9 \, \text{m/s}$ (also approaching, hence negative).Substitute the values into the differentiation formula: $\frac{d(\text{Distance})}{dt} = (-2 \times \frac{8}{10}) + (-9 \times \frac{6}{10})$ .
Rate of Change Calculation: The velocities of the cars are given as $\frac{dx}{dt} = -2 \, \text{m/s}$ (since it's approaching the intersection, we take it as negative) and $\frac{dy}{dt} = -9 \, \text{m/s}$ (also approaching, hence negative).Substitute the values into the differentiation formula: $\frac{d(\text{Distance})}{dt} = (-2 \times \frac{8}{10}) + (-9 \times \frac{6}{10})$ .Calculate the rate of change: $\frac{d(\text{Distance})}{dt} = (-\frac{16}{10}) + (-\frac{54}{10})$ .