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Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 2 meters per second and the second car's velocity is 9 meters per second. At a certain instant, the first car is 8 meters from the intersection and the second car is 6 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) -sqrt85 (B) -7 (C) -8.4 (D) -10

Two cars are driving towards an intersection from perpendicular directions.\newlineThe first car's velocity is 22 meters per second and the second car's velocity is 99 meters per second.\newlineAt a certain instant, the first car is 88 meters from the intersection and the second car is 66 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) −85 -\sqrt{85} \newline(B) −7-7\newline(C) −8-8.44\newline(D) −10-10

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Q. Two cars are driving towards an intersection from perpendicular directions.\newlineThe first car's velocity is 22 meters per second and the second car's velocity is 99 meters per second.\newlineAt a certain instant, the first car is 88 meters from the intersection and the second car is 66 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) −85 -\sqrt{85} \newline(B) −7-7\newline(C) −8-8.44\newline(D) −10-10
  1. Define Distances: Let's call the distance between the first car and the intersection x1x_1, and the distance between the second car and the intersection x2x_2. We know that x1=8x_1 = 8 meters and x2=6x_2 = 6 meters.
  2. Apply Pythagorean Theorem: The distance between the two cars can be found using the Pythagorean theorem, since they are approaching the intersection from perpendicular directions. Let's call this distance dd.\newlineSo, d2=x12+x22d^2 = x_1^2 + x_2^2.
  3. Calculate Distance: Plugging in the values, we get d2=82+62=64+36=100d^2 = 8^2 + 6^2 = 64 + 36 = 100.
  4. Find Rate of Change: Taking the square root of both sides to find dd, we get d=100=10d = \sqrt{100} = 10 meters.
  5. Derive Rate of Change: Now, we need to find the rate of change of this distance with respect to time. We can use the derivative of the distance function with respect to time, which is the sum of the derivatives of x1x_1 and x2x_2 with respect to time, since they are both functions of time.
  6. Derive Rate of Change: Now, we need to find the rate of change of this distance with respect to time. We can use the derivative of the distance function with respect to time, which is the sum of the derivatives of x1x_1 and x2x_2 with respect to time, since they are both functions of time.The rate of change of x1x_1 with respect to time is the velocity of the first car, which is −2-2 meters per second (negative because it's approaching the intersection). Similarly, the rate of change of x2x_2 with respect to time is the velocity of the second car, which is −9-9 meters per second.
  7. Derive Rate of Change: Now, we need to find the rate of change of this distance with respect to time. We can use the derivative of the distance function with respect to time, which is the sum of the derivatives of x1x_1 and x2x_2 with respect to time, since they are both functions of time.The rate of change of x1x_1 with respect to time is the velocity of the first car, which is −2-2 meters per second (negative because it's approaching the intersection). Similarly, the rate of change of x2x_2 with respect to time is the velocity of the second car, which is −9-9 meters per second.The rate of change of the distance dd with respect to time is the derivative of the square root of (x12+x22)(x_1^2 + x_2^2) with respect to time. Using the chain rule, we get (1/2)(2x1(dx1/dt)+2x2(dx2/dt))/x12+x22(1/2)(2x_1(dx_1/dt) + 2x_2(dx_2/dt)) / \sqrt{x_1^2 + x_2^2}.
  8. Derive Rate of Change: Now, we need to find the rate of change of this distance with respect to time. We can use the derivative of the distance function with respect to time, which is the sum of the derivatives of x1x_1 and x2x_2 with respect to time, since they are both functions of time.The rate of change of x1x_1 with respect to time is the velocity of the first car, which is −2-2 meters per second (negative because it's approaching the intersection). Similarly, the rate of change of x2x_2 with respect to time is the velocity of the second car, which is −9-9 meters per second.The rate of change of the distance dd with respect to time is the derivative of the square root of (x12+x22)(x_1^2 + x_2^2) with respect to time. Using the chain rule, we get (1/2)(2x1(dx1/dt)+2x2(dx2/dt))/x12+x22(1/2)(2x_1(dx_1/dt) + 2x_2(dx_2/dt)) / \sqrt{x_1^2 + x_2^2}.Plugging in the values, we get (1/2)(2⋅8⋅(−2)+2⋅6⋅(−9))/64+36(1/2)(2\cdot8\cdot(-2) + 2\cdot6\cdot(-9)) / \sqrt{64 + 36}.
  9. Derive Rate of Change: Now, we need to find the rate of change of this distance with respect to time. We can use the derivative of the distance function with respect to time, which is the sum of the derivatives of x1x_1 and x2x_2 with respect to time, since they are both functions of time.The rate of change of x1x_1 with respect to time is the velocity of the first car, which is −2-2 meters per second (negative because it's approaching the intersection). Similarly, the rate of change of x2x_2 with respect to time is the velocity of the second car, which is −9-9 meters per second.The rate of change of the distance dd with respect to time is the derivative of the square root of (x12+x22)(x_1^2 + x_2^2) with respect to time. Using the chain rule, we get (1/2)(2x1(dx1/dt)+2x2(dx2/dt))/x12+x22(1/2)(2x_1(dx_1/dt) + 2x_2(dx_2/dt)) / \sqrt{x_1^2 + x_2^2}.Plugging in the values, we get (1/2)(2⋅8⋅(−2)+2⋅6⋅(−9))/64+36(1/2)(2\cdot8\cdot(-2) + 2\cdot6\cdot(-9)) / \sqrt{64 + 36}.Simplifying, we get x2x_200 meters per second.

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