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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) -8.4 (B) -7 (C) -10 (D) -sqrt85

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) −8-8.44\newline(B) −7-7\newline(C) −10-10\newline(D) −85 -\sqrt{85}

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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) −8-8.44\newline(B) −7-7\newline(C) −10-10\newline(D) −85 -\sqrt{85}
  1. Initial Distance Calculation: First car's distance from intersection is 88 meters, second car's distance is 66 meters. We need to find the rate at which the distance between them is changing.
  2. Pythagoras' Theorem Application: Let's use Pythagoras' theorem to find the initial distance between the cars. Distance2=82+62\text{Distance}^2 = 8^2 + 6^2.
  3. Calculation of Initial Distance: Calculating the initial distance: Distance2=64+36\text{Distance}^2 = 64 + 36.
  4. Differentiation for Rate of Change: Distance2=100\text{Distance}^2 = 100, so Distance=100\text{Distance} = \sqrt{100}.
  5. Velocities of the Cars: Distance = 1010 meters. This is the initial distance between the cars.
  6. Substitution of Values: Now, let's differentiate the distance with respect to time to find the rate of change. If xx is the distance of the first car and yy is the distance of the second car from the intersection, then using the chain rule, d(Distance)dt=(dxdt∗xDistance)+(dydt∗yDistance)\frac{d(\text{Distance})}{dt} = \left(\frac{dx}{dt} * \frac{x}{\text{Distance}}\right) + \left(\frac{dy}{dt} * \frac{y}{\text{Distance}}\right).
  7. Rate of Change Calculation: The velocities of the cars are given as dxdt=−2 m/s\frac{dx}{dt} = -2 \, \text{m/s} (since it's approaching the intersection, we take it as negative) and dydt=−9 m/s\frac{dy}{dt} = -9 \, \text{m/s} (also approaching, hence negative).
  8. Rate of Change Calculation: The velocities of the cars are given as dxdt=−2 m/s\frac{dx}{dt} = -2 \, \text{m/s} (since it's approaching the intersection, we take it as negative) and dydt=−9 m/s\frac{dy}{dt} = -9 \, \text{m/s} (also approaching, hence negative).Substitute the values into the differentiation formula: d(Distance)dt=(−2×810)+(−9×610)\frac{d(\text{Distance})}{dt} = (-2 \times \frac{8}{10}) + (-9 \times \frac{6}{10}).
  9. Rate of Change Calculation: The velocities of the cars are given as dxdt=−2 m/s\frac{dx}{dt} = -2 \, \text{m/s} (since it's approaching the intersection, we take it as negative) and dydt=−9 m/s\frac{dy}{dt} = -9 \, \text{m/s} (also approaching, hence negative).Substitute the values into the differentiation formula: d(Distance)dt=(−2×810)+(−9×610)\frac{d(\text{Distance})}{dt} = (-2 \times \frac{8}{10}) + (-9 \times \frac{6}{10}).Calculate the rate of change: d(Distance)dt=(−1610)+(−5410)\frac{d(\text{Distance})}{dt} = (-\frac{16}{10}) + (-\frac{54}{10}).

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