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) $25$ $\newline$ (C) $9$ . $4$ $\newline$ (D) $\sqrt{89}$

View step-by-step help

Home

Math Problems

Grade 8

Solve equations with variables on both sides: word problems

Full solution

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)

25

\newline

(C)

9

4

\newline

(D)

\sqrt{89}

Formulating Triangle Hypotenuse: The distance between the two cars can be represented by the hypotenuse of a right triangle, where the legs are the distances of each car from the intersection.
Calculating Initial Distance: Let's call the distance of the first car from the intersection $x_1$ ( $15$ meters) and the second car $x_2$ ( $20$ meters). The hypotenuse (distance between the cars) is $d$ .
Finding Square Root: Using Pythagoras' theorem, we calculate the initial distance between the cars: $d^2 = x_1^2 + x_2^2$ .
Determining Rates of Change: Plugging in the values: $d^2 = 15^2 + 20^2 = 225 + 400 = 625$ .
Applying Chain Rule: Taking the square root to find $d$ : $d = \sqrt{625} = 25$ meters.
Solving for $\frac{d}{dt}$ : Now, we need to find the rate of change of $d$ with respect to time ( $\frac{d}{dt}$ ). Since the cars are moving away from the intersection, we differentiate both $x_1$ and $x_2$ with respect to time.
Solving for $\frac{d}{dt}$ : Now, we need to find the rate of change of $d$ with respect to time ( $\frac{d}{dt}$ ). Since the cars are moving away from the intersection, we differentiate both $x_1$ and $x_2$ with respect to time.The rate of change of $x_1$ with respect to time ( $\frac{dx_1}{dt}$ ) is the velocity of the first car, which is $5$ m/s. Similarly, the rate of change of $x_2$ with respect to time ( $\frac{dx_2}{dt}$ ) is the velocity of the second car, which is $d$ $0$ m/s.
Solving for $\frac{d}{dt}$ : Now, we need to find the rate of change of $d$ with respect to time ( $\frac{dd}{dt}$ ). Since the cars are moving away from the intersection, we differentiate both $x_1$ and $x_2$ with respect to time.The rate of change of $x_1$ with respect to time ( $\frac{dx_1}{dt}$ ) is the velocity of the first car, which is $5$ m/s. Similarly, the rate of change of $x_2$ with respect to time ( $\frac{dx_2}{dt}$ ) is the velocity of the second car, which is $d$ $0$ m/s.Using the chain rule for differentiation, we get: $d$ $1$ .
Solving for $\frac{d}{dt}$ : Now, we need to find the rate of change of $d$ with respect to time ( $\frac{d}{dt}$ ). Since the cars are moving away from the intersection, we differentiate both $x_1$ and $x_2$ with respect to time.The rate of change of $x_1$ with respect to time ( $\frac{dx_1}{dt}$ ) is the velocity of the first car, which is $5$ m/s. Similarly, the rate of change of $x_2$ with respect to time ( $\frac{dx_2}{dt}$ ) is the velocity of the second car, which is $d$ $0$ m/s.Using the chain rule for differentiation, we get: $d$ $1$ .Plugging in the values: $d$ $2$ .
Solving for $\frac{d}{dt}$ : Now, we need to find the rate of change of $d$ with respect to time ( $\frac{d}{dt}$ ). Since the cars are moving away from the intersection, we differentiate both $x_1$ and $x_2$ with respect to time.The rate of change of $x_1$ with respect to time ( $\frac{dx_1}{dt}$ ) is the velocity of the first car, which is $5$ m/s. Similarly, the rate of change of $x_2$ with respect to time ( $\frac{dx_2}{dt}$ ) is the velocity of the second car, which is $d$ $0$ m/s.Using the chain rule for differentiation, we get: $d$ $1$ .Plugging in the values: $d$ $2$ .Simplify the equation: $d$ $3$ .
Solving for $\frac{d}{dt}$ : Now, we need to find the rate of change of $d$ with respect to time ( $\frac{d}{dt}$ ). Since the cars are moving away from the intersection, we differentiate both $x_1$ and $x_2$ with respect to time.The rate of change of $x_1$ with respect to time ( $\frac{dx_1}{dt}$ ) is the velocity of the first car, which is $5$ m/s. Similarly, the rate of change of $x_2$ with respect to time ( $\frac{dx_2}{dt}$ ) is the velocity of the second car, which is $d$ $0$ m/s.Using the chain rule for differentiation, we get: $d$ $1$ .Plugging in the values: $d$ $2$ .Simplify the equation: $d$ $3$ .Divide both sides by $d$ $4$ to solve for ( $\frac{d}{dt}$ ): $d$ $6$ m/s.