A person stands $10$ meters east of an intersection and watches a car driving towards the intersection from the north at $13$ meters per second. $\newline$ At a certain instant, the car is $24$ meters from the intersection. $\newline$ What is the rate of change of the distance between the car and the person at that instant (in meters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $-26$ $\newline$ (B) $-\sqrt{269}$ $\newline$ (C) $-12$ $\newline$ (D) $-\frac{169}{12}$

View step-by-step help

Home

Math Problems

Algebra 1

Rate of travel: word problems

Full solution

Q. A person stands

10

meters east of an intersection and watches a car driving towards the intersection from the north at

13

meters per second.

\newline

At a certain instant, the car is

24

meters from the intersection.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-26

\newline

(B)

-\sqrt{269}

\newline

(C)

-12

\newline

(D)

-\frac{169}{12}

Draw Triangle: First, let's draw a right triangle where the car's distance from the intersection is one leg ( $24$ meters), and the person's distance from the intersection is the other leg ( $10$ meters). We're looking for the rate of change of the hypotenuse, which represents the distance between the car and the person.
Calculate Initial Distance: Using the Pythagorean theorem, we calculate the initial distance between the car and the person. $\newline$ $\text{Distance}^2 = 24^2 + 10^2$ $\newline$ $\text{Distance}^2 = 576 + 100$ $\newline$ $\text{Distance}^2 = 676$ $\newline$ $\text{Distance} = \sqrt{676}$ $\newline$ $\text{Distance} = 26 \text{ meters}$
Find Rate of Change: Now, we need to find the rate of change of this distance. Since the car is moving towards the intersection, the distance between the car and the person is decreasing. We'll use related rates to find this rate of change.
Use Related Rates: Let's denote the distance between the car and the intersection as $x$ , the distance between the person and the intersection as $y$ (which is constant at $10$ meters), and the distance between the car and the person as $z$ .
Simplify Equation: The Pythagorean theorem in terms of $x$ , $y$ , and $z$ is: $x^2 + y^2 = z^2$ Differentiating both sides with respect to time $t$ , we get: $2x\left(\frac{dx}{dt}\right) + 2y\left(\frac{dy}{dt}\right) = 2z\left(\frac{dz}{dt}\right)$
Apply Values: Since $y$ is constant, $\frac{dy}{dt}$ is $0$ , and we can simplify the equation to: $\newline$ $2x\frac{dx}{dt} = 2z\frac{dz}{dt}$
Calculate Rate of Change: We know that $\frac{dx}{dt}$ (the speed of the car) is $-13$ meters per second (negative because $x$ is decreasing), and we want to find $\frac{dz}{dt}$ (the rate of change of the distance between the car and the person).
Calculate Rate of Change: We know that $\frac{dx}{dt}$ (the speed of the car) is $-13$ meters per second (negative because $x$ is decreasing), and we want to find $\frac{dz}{dt}$ (the rate of change of the distance between the car and the person).Plugging in the values we have: $\newline$ $2 \times 24 \times (-13) = 2 \times 26 \times (\frac{dz}{dt})$ $\newline$ $-624 = 52 \times (\frac{dz}{dt})$ $\newline$ $(\frac{dz}{dt}) = \frac{-624}{52}$ $\newline$ $(\frac{dz}{dt}) = -12$ meters per second