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Alice is driving north at 80km/h and Bob is driving east at 50km/h. Both cars are approaching the intersection of the two roads. At what rate is the distance between the two cars decreasing when Alice's car is 0.5km from the intersection and Bob's car is 0.8km from the intersection?

Alice is driving north at 80 km/h 80 \mathrm{~km} / \mathrm{h} and Bob is driving east at 50 km/h 50 \mathrm{~km} / \mathrm{h} . Both cars are approaching the intersection of the two roads. At what rate is the distance between the two cars decreasing when Alice's car is 0.5 km 0.5 \mathrm{~km} from the intersection and Bob's car is 0.8 km 0.8 \mathrm{~km} from the intersection?

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Full solution

Q. Alice is driving north at 80 km/h 80 \mathrm{~km} / \mathrm{h} and Bob is driving east at 50 km/h 50 \mathrm{~km} / \mathrm{h} . Both cars are approaching the intersection of the two roads. At what rate is the distance between the two cars decreasing when Alice's car is 0.5 km 0.5 \mathrm{~km} from the intersection and Bob's car is 0.8 km 0.8 \mathrm{~km} from the intersection?
  1. Identify scenario and variables: Identify the scenario and the known variables.\newlineAlice's speed va=80v_a = 80 km/h northward.\newlineBob's speed vb=50v_b = 50 km/h eastward.\newlineAlice's distance from the intersection da=0.5d_a = 0.5 km.\newlineBob's distance from the intersection db=0.8d_b = 0.8 km.\newlineWe need to find the rate at which the distance between the two cars is decreasing.
  2. Visualize as right triangle: Visualize the scenario as a right triangle where the cars' paths are the legs and the distance between the cars is the hypotenuse. The rate at which Alice is approaching the intersection affects the vertical leg, and the rate at which Bob is approaching affects the horizontal leg.
  3. Apply Pythagorean theorem: Apply the Pythagorean theorem to find the initial distance between the two cars.\newlineLet dd be the distance between the cars.\newlined2=da2+db2d^2 = d_a^2 + d_b^2\newlined2=(0.5)2+(0.8)2d^2 = (0.5)^2 + (0.8)^2\newlined2=0.25+0.64d^2 = 0.25 + 0.64\newlined2=0.89d^2 = 0.89\newlined=0.89d = \sqrt{0.89}\newlined≈0.943d \approx 0.943 km
  4. Use related rates concept: Use the related rates concept from calculus, where the rates at which the sides of the triangle change are related to the rate at which the hypotenuse changes.\newlineLet's denote the rate at which the distance between the cars is decreasing as ddt\frac{d}{dt}.\newlineUsing the chain rule, we have:\newline2d(ddt)=2da(dadt)+2db(dbdt)2d\left(\frac{d}{dt}\right) = 2d_a\left(\frac{da}{dt}\right) + 2d_b\left(\frac{db}{dt}\right)\newlineSince Alice is moving towards the intersection, dadt=−va\frac{da}{dt} = -v_a, and since Bob is moving towards the intersection, dbdt=−vb\frac{db}{dt} = -v_b.
  5. Plug in values and solve: Plug in the known values and solve for dddt\frac{dd}{dt}.
    2×0.943×(dddt)=2×0.5×(−80)+2×0.8×(−50)2 \times 0.943 \times \left(\frac{dd}{dt}\right) = 2 \times 0.5 \times (-80) + 2 \times 0.8 \times (-50)
    1.886×(dddt)=−80+(−80)1.886 \times \left(\frac{dd}{dt}\right) = -80 + (-80)
    1.886×(dddt)=−1601.886 \times \left(\frac{dd}{dt}\right) = -160
    dddt=−1601.886\frac{dd}{dt} = \frac{-160}{1.886}
    dddt≈−84.83 km/h\frac{dd}{dt} \approx -84.83 \text{ km/h}

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