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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) 25 (C) 9.4 (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) 2525\newline(C) 99.44\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) 2525\newline(C) 99.44\newline(D) 89 \sqrt{89}
  1. 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.
  2. Calculating Initial Distance: Let's call the distance of the first car from the intersection x1x_1 (1515 meters) and the second car x2x_2 (2020 meters). The hypotenuse (distance between the cars) is dd.
  3. Finding Square Root: Using Pythagoras' theorem, we calculate the initial distance between the cars: d2=x12+x22d^2 = x_1^2 + x_2^2.
  4. Determining Rates of Change: Plugging in the values: d2=152+202=225+400=625d^2 = 15^2 + 20^2 = 225 + 400 = 625.
  5. Applying Chain Rule: Taking the square root to find dd: d=625=25d = \sqrt{625} = 25 meters.
  6. Solving for ddt\frac{d}{dt}: Now, we need to find the rate of change of dd with respect to time (ddt\frac{d}{dt}). Since the cars are moving away from the intersection, we differentiate both x1x_1 and x2x_2 with respect to time.
  7. Solving for ddt\frac{d}{dt}: Now, we need to find the rate of change of dd with respect to time (ddt\frac{d}{dt}). Since the cars are moving away from the intersection, we differentiate both x1x_1 and x2x_2 with respect to time.The rate of change of x1x_1 with respect to time (dx1dt\frac{dx_1}{dt}) is the velocity of the first car, which is 55 m/s. Similarly, the rate of change of x2x_2 with respect to time (dx2dt\frac{dx_2}{dt}) is the velocity of the second car, which is dd00 m/s.
  8. Solving for ddt\frac{d}{dt}: Now, we need to find the rate of change of dd with respect to time (dddt\frac{dd}{dt}). Since the cars are moving away from the intersection, we differentiate both x1x_1 and x2x_2 with respect to time.The rate of change of x1x_1 with respect to time (dx1dt\frac{dx_1}{dt}) is the velocity of the first car, which is 55 m/s. Similarly, the rate of change of x2x_2 with respect to time (dx2dt\frac{dx_2}{dt}) is the velocity of the second car, which is dd00 m/s.Using the chain rule for differentiation, we get: dd11.
  9. Solving for ddt\frac{d}{dt}: Now, we need to find the rate of change of dd with respect to time (ddt\frac{d}{dt}). Since the cars are moving away from the intersection, we differentiate both x1x_1 and x2x_2 with respect to time.The rate of change of x1x_1 with respect to time (dx1dt\frac{dx_1}{dt}) is the velocity of the first car, which is 55 m/s. Similarly, the rate of change of x2x_2 with respect to time (dx2dt\frac{dx_2}{dt}) is the velocity of the second car, which is dd00 m/s.Using the chain rule for differentiation, we get: dd11.Plugging in the values: dd22.
  10. Solving for ddt\frac{d}{dt}: Now, we need to find the rate of change of dd with respect to time (ddt\frac{d}{dt}). Since the cars are moving away from the intersection, we differentiate both x1x_1 and x2x_2 with respect to time.The rate of change of x1x_1 with respect to time (dx1dt\frac{dx_1}{dt}) is the velocity of the first car, which is 55 m/s. Similarly, the rate of change of x2x_2 with respect to time (dx2dt\frac{dx_2}{dt}) is the velocity of the second car, which is dd00 m/s.Using the chain rule for differentiation, we get: dd11.Plugging in the values: dd22.Simplify the equation: dd33.
  11. Solving for ddt\frac{d}{dt}: Now, we need to find the rate of change of dd with respect to time (ddt\frac{d}{dt}). Since the cars are moving away from the intersection, we differentiate both x1x_1 and x2x_2 with respect to time.The rate of change of x1x_1 with respect to time (dx1dt\frac{dx_1}{dt}) is the velocity of the first car, which is 55 m/s. Similarly, the rate of change of x2x_2 with respect to time (dx2dt\frac{dx_2}{dt}) is the velocity of the second car, which is dd00 m/s.Using the chain rule for differentiation, we get: dd11.Plugging in the values: dd22.Simplify the equation: dd33.Divide both sides by dd44 to solve for (ddt\frac{d}{dt}): dd66 m/s.

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