A person stands $30$ meters east of an intersection and watches a car driving away from the intersection to the north at $17$ meters per second. $\newline$ At a certain instant, the car is $16$ 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) $34$ $\newline$ (B) $8$ $\newline$ (C) $36$ . $125$ $\newline$ (D) $\sqrt{1189}$

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Grade 7

Calculate unit rates with fractions

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Q. A person stands

30

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

17

meters per second.

\newline

At a certain instant, the car is

16

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)

34

\newline

(B)

8

\newline

(C)

36

125

\newline

(D)

\sqrt{1189}

Identify Triangle Type: We're dealing with a right triangle where the car is one leg ( $16$ meters) and the person is the other leg ( $30$ meters). We need to find the rate of change of the hypotenuse (distance between the car and the person) when the car is $16$ meters from the intersection.
Calculate Initial Distance: First, let's find the initial distance between the car and the person using the Pythagorean theorem: $\text{distance}^2 = 30^2 + 16^2$ .
Find Initial Distance: Calculating the initial distance: $\text{distance}^2 = 900 + 256 = 1156$ .
Apply Related Rates: Taking the square root to find the distance: $\text{distance} = \sqrt{1156} = 34 \text{ meters}.$
Use Pythagorean Theorem: Now, we need to use related rates to find the rate of change of the distance. Let's call the distance between the car and the person ' $d$ ', the distance of the car from the intersection ' $x$ ', and the rate of change of ' $d$ ' as ' $\frac{dd}{dt}$ '. We know ' $\frac{dx}{dt}$ ' (the speed of the car) is $17$ meters per second.
Use Pythagorean Theorem: Now, we need to use related rates to find the rate of change of the distance. Let's call the distance between the car and the person 'd ext{', the distance of the car from the intersection '}}x ext{', and the rate of change of '}}d ext{' as '}}rac{dd}{dt} ext{'. We know '}}rac{dx}{dt} ext{' (the speed of the car) is }} ext{\(17 meters per second.Using related rates, we have }} ext{(}}rac{dd}{dt} ext{)}^ $2$ = ext{(}}rac{dx}{dt} ext{)}^ $2$ + ext{(}}rac{dy}{dt} ext{)}^ $2$ ext{, where '}}dy/dt ext{' is the rate of change of the person's distance from the intersection, which is }} ext{ $0$ since the person is standing still.}
Use Pythagorean Theorem: Now, we need to use related rates to find the rate of change of the distance. Let's call the distance between the car and the person 'd ext{', the distance of the car from the intersection '} extit{x}', and the rate of change of ' extit{d}' as 'rac{dd}{dt}'. We know 'rac{dx}{dt}' (the speed of the car) is extit{$17$} meters per second.Using related rates, we have extit{igg(rac{dd}{dt}igg)^$2$ = igg(rac{dx}{dt}igg)^$2$ + igg(rac{dy}{dt}igg)^$2$}, where 'rac{dy}{dt}' is the rate of change of the person's distance from the intersection, which is extit{$0$} since the person is standing still.Substituting the known values, we get extit{igg(rac{dd}{dt}igg)^$2$ = $17$^$2$ + $0$^$2$}.
Use Pythagorean Theorem: Now, we need to use related rates to find the rate of change of the distance. Let's call the distance between the car and the person '\(d', the distance of the car from the intersection ' $x$ ', and the rate of change of ' $d$ ' as ' $\frac{dd}{dt}$ '. We know ' $\frac{dx}{dt}$ ' (the speed of the car) is $17$ meters per second.Using related rates, we have $(\frac{dd}{dt})^2 = (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2$ , where ' $\frac{dy}{dt}$ ' is the rate of change of the person's distance from the intersection, which is $0$ since the person is standing still.Substituting the known values, we get $(\frac{dd}{dt})^2 = 17^2 + 0^2$ .Calculating the rate of change of the distance: $x$ $0$ .
Use Pythagorean Theorem: Now, we need to use related rates to find the rate of change of the distance. Let's call the distance between the car and the person 'd ext{', the distance of the car from the intersection '} extit{x} ext{', and the rate of change of '} extit{d} ext{' as '}rac{dd}{dt} ext{'. We know '}rac{dx}{dt} ext{' (the speed of the car) is }\(17 ext{ meters per second.Using related rates, we have } ext{(}rac{dd}{dt} ext{)}^ $2$ = ext{(}rac{dx}{dt} ext{)}^ $2$ + ext{(}rac{dy}{dt} ext{)}^ $2$ ext{, where '}rac{dy}{dt} ext{' is the rate of change of the person's distance from the intersection, which is } $0$ ext{ since the person is standing still.Substituting the known values, we get } ext{(}rac{dd}{dt} ext{)}^ $2$ = $17$ ^ $2$ + $0$ ^ $2$ ext{.Calculating the rate of change of the distance: } ext{(}rac{dd}{dt} ext{)}^ $2$ = $289$ ext{.Taking the square root to find }rac{dd}{dt} ext{: }rac{dd}{dt} = ext{sqrt}( $289$ ) = $17$ ext{ meters per second. But this is incorrect because we didn't consider the direction of the car's movement relative to the person. The car is moving away, so the distance is increasing. We need to use the Pythagorean theorem to find the correct rate of change.}