Two cars are driving away from an intersection in perpendicular directions. $\newline$ The first car's velocity is $7$ meters per second and the second car's velocity is $3$ meters per second. $\newline$ At a certain instant, the first car is $5$ meters from the intersection and the second car is $12$ 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) $\frac{99}{13}$ $\newline$ (B) $\frac{71}{13}$ $\newline$ (C) $\mathbf{1 3}$ $\newline$ (D) $\sqrt{58}$

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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

7

meters per second and the second car's velocity is

3

meters per second.

\newline

At a certain instant, the first car is

5

meters from the intersection and the second car is

12

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)

\frac{99}{13}

\newline

(B)

\frac{71}{13}

\newline

(C)

\mathbf{1 3}

\newline

(D)

\sqrt{58}

Pythagorean Theorem Application: The cars are moving away from the intersection in perpendicular directions, so we can use the Pythagorean theorem to find the distance between them.
Calculation of Distance: Let's call the distance between the cars ＂d＂. At the instant we're considering, the first car is $5$ meters from the intersection and the second car is $12$ meters away. So, $d^2 = 5^2 + 12^2$ .
Finding Rate of Change: Calculating $d^2$ gives us $d^2 = 25 + 144$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters.
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters. Now we need to find the rate of change of $d$ with respect to time, which is denoted as $\frac{dd}{dt}$ . We can use the derivative of the Pythagorean theorem for this, which is $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2d\frac{dd}{dt}$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters. Now we need to find the rate of change of $d$ with respect to time, which is denoted as $\frac{dd}{dt}$ . We can use the derivative of the Pythagorean theorem for this, which is $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2d\frac{dd}{dt}$ . Substituting the given velocities for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , we have $2\cdot5\cdot7 + 2\cdot12\cdot3 = 2\cdot13\cdot\frac{dd}{dt}$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters. Now we need to find the rate of change of $d$ with respect to time, which is denoted as $\frac{dd}{dt}$ . We can use the derivative of the Pythagorean theorem for this, which is $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2d\frac{dd}{dt}$ . Substituting the given velocities for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , we have $2\cdot5\cdot7 + 2\cdot12\cdot3 = 2\cdot13\cdot\frac{dd}{dt}$ . Calculating the left side, we get $d$ $0$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters. Now we need to find the rate of change of $d$ with respect to time, which is denoted as $\frac{dd}{dt}$ . We can use the derivative of the Pythagorean theorem for this, which is $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2d\frac{dd}{dt}$ . Substituting the given velocities for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , we have $2\cdot5\cdot7 + 2\cdot12\cdot3 = 2\cdot13\cdot\frac{dd}{dt}$ . Calculating the left side, we get $d$ $0$ . Adding $d$ $1$ and $d$ $2$ gives us $d$ $3$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters. Now we need to find the rate of change of $d$ with respect to time, which is denoted as $\frac{dd}{dt}$ . We can use the derivative of the Pythagorean theorem for this, which is $2x(\frac{dx}{dt}) + 2y(\frac{dy}{dt}) = 2d(\frac{dd}{dt})$ . Substituting the given velocities for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , we have $2*5*7 + 2*12*3 = 2*13*(\frac{dd}{dt})$ . Calculating the left side, we get $d$ $0$ . Adding $d$ $1$ and $d$ $2$ gives us $d$ $3$ . Dividing both sides by $d$ $4$ to solve for $\frac{dd}{dt}$ , we get $d$ $6$ .
Derivative Calculation: Adding those up, we get $d^2 = 169$ . Taking the square root of both sides to find $d$ , we get $d = \sqrt{169}$ . So, $d = 13$ meters. Now we need to find the rate of change of $d$ with respect to time, which is denoted as $\frac{dd}{dt}$ . We can use the derivative of the Pythagorean theorem for this, which is $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2d\frac{dd}{dt}$ . Substituting the given velocities for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , we have $2\cdot5\cdot7 + 2\cdot12\cdot3 = 2\cdot13\cdot\frac{dd}{dt}$ . Calculating the left side, we get $d$ $0$ . Adding $d$ $1$ and $d$ $2$ gives us $d$ $3$ . Dividing both sides by $d$ $4$ to solve for $\frac{dd}{dt}$ , we get $d$ $6$ . Simplifying $d$ $7$ gives us $d$ $8$ meters per second.