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Two cars are driving towards an intersection from perpendicular directions.
The first car's velocity is 10 meters per second and the second car's velocity is 6 meters per second.
At a certain instant, the first car is 4 meters from the intersection and the second car is 3 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) -11.6
(B) -10.8
(C) -5
(D) 
-sqrt136

Two cars are driving towards an intersection from perpendicular directions.\newlineThe first car's velocity is 1010 meters per second and the second car's velocity is 66 meters per second.\newlineAt a certain instant, the first car is 44 meters from the intersection and the second car is 33 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) 11-11.66\newline(B) 10-10.88\newline(C) 5-5\newline(D) 136 -\sqrt{136}

Full solution

Q. Two cars are driving towards an intersection from perpendicular directions.\newlineThe first car's velocity is 1010 meters per second and the second car's velocity is 66 meters per second.\newlineAt a certain instant, the first car is 44 meters from the intersection and the second car is 33 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) 11-11.66\newline(B) 10-10.88\newline(C) 5-5\newline(D) 136 -\sqrt{136}
  1. Formulating Right Triangle: The distance between the cars can be represented by the hypotenuse of a right triangle, with the sides being the distances of the cars from the intersection.
  2. Applying Pythagorean Theorem: Let's call the distance between the cars "d". Using the Pythagorean theorem, d2=42+32d^2 = 4^2 + 3^2.
  3. Calculating Distance: Calculating dd, we get d2=16+9d^2 = 16 + 9 which is d2=25d^2 = 25.
  4. Finding Rate of Change: Taking the square root of both sides, d=25d = \sqrt{25}, so d=5d = 5 meters.
  5. Using Chain Rule: Now, we need to find the rate of change of dd with respect to time, which is dddt\frac{dd}{dt}.
  6. Differentiating Distances: Using the chain rule, dddt=(dddx1)(dx1dt)+(dddx2)(dx2dt)\frac{dd}{dt} = \left(\frac{dd}{dx_1}\right)\left(\frac{dx_1}{dt}\right) + \left(\frac{dd}{dx_2}\right)\left(\frac{dx_2}{dt}\right), where x1x_1 and x2x_2 are the distances of the cars from the intersection.
  7. Determining Velocities: Differentiating dd with respect to x1x_1 and x2x_2, we get rac{dd}{dx_1} = rac{4}{5} and rac{dd}{dx_2} = rac{3}{5}.
  8. Calculating Rate of Change: The rates dx1dt\frac{dx_1}{dt} and dx2dt\frac{dx_2}{dt} are the velocities of the cars, which are 10m/s-10\,\text{m/s} and 6m/s-6\,\text{m/s}, respectively (negative because they are approaching the intersection).
  9. Calculating Rate of Change: The rates dx1dt\frac{dx_1}{dt} and dx2dt\frac{dx_2}{dt} are the velocities of the cars, which are 10m/s-10 \, \text{m/s} and 6m/s-6 \, \text{m/s}, respectively (negative because they are approaching the intersection).Plugging in the values, d2ddt2=(45)(10)+(35)(6)\frac{d^2d}{dt^2} = \left(\frac{4}{5}\right)(-10) + \left(\frac{3}{5}\right)(-6).
  10. Calculating Rate of Change: The rates dx1dt\frac{dx_1}{dt} and dx2dt\frac{dx_2}{dt} are the velocities of the cars, which are 10m/s-10 \, \text{m/s} and 6m/s-6 \, \text{m/s}, respectively (negative because they are approaching the intersection).Plugging in the values, d2ddt=(45)(10)+(35)(6)\frac{d^2d}{dt} = \left(\frac{4}{5}\right)(-10) + \left(\frac{3}{5}\right)(-6).Calculating d2ddt\frac{d^2d}{dt}, we get d2ddt=83.6\frac{d^2d}{dt} = -8 - 3.6 which is d2ddt=11.6\frac{d^2d}{dt} = -11.6 meters per second.

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