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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters per second.
At a certain instant, the car is 24 meters from the intersection.
What is the rate of change of the distance between the car and the person at that instant (in meters per second)?
Choose 1 answer:
(A) 
-sqrt269
(B) -26
(C) -12
(D) 
-(169)/(12)

A person stands 1010 meters east of an intersection and watches a car driving towards the intersection from the north at 1313 meters per second.\newlineAt a certain instant, the car is 2424 meters from the intersection.\newlineWhat is the rate of change of the distance between the car and the person at that instant (in meters per second)?\newlineChoose 11 answer:\newline(A) 269 -\sqrt{269} \newline(B) 26-26\newline(C) 12-12\newline(D) 16912 -\frac{169}{12}

Full solution

Q. A person stands 1010 meters east of an intersection and watches a car driving towards the intersection from the north at 1313 meters per second.\newlineAt a certain instant, the car is 2424 meters from the intersection.\newlineWhat is the rate of change of the distance between the car and the person at that instant (in meters per second)?\newlineChoose 11 answer:\newline(A) 269 -\sqrt{269} \newline(B) 26-26\newline(C) 12-12\newline(D) 16912 -\frac{169}{12}
  1. Draw Triangle: First, let's draw a right triangle where the car is one vertex, the person is the second vertex, and the intersection is the third vertex. The car is 2424 meters from the intersection (north side), and the person is 1010 meters from the intersection (east side).
  2. Find Hypotenuse: We need to find the hypotenuse of the triangle, which represents the distance between the car and the person at that instant. We use the Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2, where aa is 2424 meters, bb is 1010 meters, and cc is the hypotenuse.
  3. Calculate Hypotenuse: Calculating the hypotenuse: c2=242+102=576+100=676c^2 = 24^2 + 10^2 = 576 + 100 = 676, so c=676=26c = \sqrt{676} = 26 meters.
  4. Find Rate of Change: Now, we need to find the rate of change of this distance. Since the car is moving towards the intersection, the distance between the car and the person is decreasing. We'll use related rates from calculus. Let's call the distance between the car and the person 'zz', the distance of the car from the intersection 'xx', and the distance of the person from the intersection 'yy'. We have dzdt=(dxdtxz)+(dydtyz)\frac{dz}{dt} = \left(\frac{dx}{dt} \cdot \frac{x}{z}\right) + \left(\frac{dy}{dt} \cdot \frac{y}{z}\right). Since the person is not moving, dydt=0\frac{dy}{dt} = 0.
  5. Use Related Rates: We know dxdt=13\frac{dx}{dt} = -13 meters per second (negative because the distance xx is decreasing), x=24x = 24 meters, y=10y = 10 meters, and z=26z = 26 meters. Plugging these into the formula: dzdt=(13×2426)+(0×1026)\frac{dz}{dt} = \left(-13 \times \frac{24}{26}\right) + \left(0 \times \frac{10}{26}\right).
  6. Apply Calculus: Simplifying the equation: dzdt=31226+0=12\frac{dz}{dt} = \frac{-312}{26} + 0 = -12 meters per second. So, the rate of change of the distance between the car and the person is 12-12 meters per second.

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