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Two cars are driving away from an intersection in perpendicular directions. The first car's velocity is 5 meters per second and the second car's velocity is 8 meters per second. At a certain instant, the first car is 15 meters from the intersection and the second car is 20 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer: (A) 13 (B) 9.4 (C) 25 (D) sqrt89

Two cars are driving away from an intersection in perpendicular directions.\newlineThe first car's velocity is 55 meters per second and the second car's velocity is 88 meters per second.\newlineAt a certain instant, the first car is 1515 meters from the intersection and the second car is 2020 meters from the intersection.\newlineWhat is the rate of change of the distance between the cars at that instant (in meters per second)?\newlineChoose 11 answer:\newline(A) 1313\newline(B) 99.44\newline(C) 2525\newline(D) 89 \sqrt{89}

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Q. Two cars are driving away from an intersection in perpendicular directions.\newlineThe first car's velocity is 55 meters per second and the second car's velocity is 88 meters per second.\newlineAt a certain instant, the first car is 1515 meters from the intersection and the second car is 2020 meters from the intersection.\newlineWhat is the rate of change of the distance between the cars at that instant (in meters per second)?\newlineChoose 11 answer:\newline(A) 1313\newline(B) 99.44\newline(C) 2525\newline(D) 89 \sqrt{89}
  1. Calculate Distance: Let's call the distance between the cars 'dd'. We can use the Pythagorean theorem to find 'dd'.\newlined2=152+202d^2 = 15^2 + 20^2\newlined2=225+400d^2 = 225 + 400\newlined2=625d^2 = 625\newlined=625d = \sqrt{625}\newlined=25 meters.d = 25 \text{ meters}.
  2. Find Rate of Change: Now, we need to find the rate of change of dd with respect to time, which is dddt\frac{dd}{dt}. We can use the chain rule where dddt=dddx1dx1dt+dddx2dx2dt\frac{dd}{dt} = \frac{dd}{dx_1}\frac{dx_1}{dt} + \frac{dd}{dx_2}\frac{dx_2}{dt}, where x1x_1 and x2x_2 are the distances of the first and second car from the intersection, respectively.
  3. Apply Chain Rule: The rates dx1dt\frac{dx_1}{dt} and dx2dt\frac{dx_2}{dt} are the velocities of the cars, which are 5 m/s5\,\text{m/s} and 8 m/s8\,\text{m/s}, respectively.\newlineSo, we need to differentiate 'dd' with respect to 'x1x_1' and 'x2x_2'.\newlineUsing the Pythagorean theorem, we have d=x12+x22d = \sqrt{x_1^2 + x_2^2}.\newlineDifferentiating both sides with respect to 'x1x_1', we get dddx1=x1d\frac{dd}{dx_1} = \frac{x_1}{d}.\newlineSimilarly, dx2dt\frac{dx_2}{dt}00.
  4. Use Chain Rule Equation: Now we plug in the values of x1x_1, x2x_2, dx1dt\frac{dx_1}{dt}, and dx2dt\frac{dx_2}{dt} into the chain rule equation.\newlineddt=(x1d)(dx1dt)+(x2d)(dx2dt)\frac{d}{dt} = \left(\frac{x_1}{d}\right)\left(\frac{dx_1}{dt}\right) + \left(\frac{x_2}{d}\right)\left(\frac{dx_2}{dt}\right)\newlineddt=(1525)(5)+(2025)(8)\frac{d}{dt} = \left(\frac{15}{25}\right)(5) + \left(\frac{20}{25}\right)(8)\newlineddt=(35)(5)+(45)(8)\frac{d}{dt} = \left(\frac{3}{5}\right)(5) + \left(\frac{4}{5}\right)(8)\newlineddt=3+6.4\frac{d}{dt} = 3 + 6.4\newlineddt=9.4\frac{d}{dt} = 9.4 meters per second.

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