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AP Calculus AB· difficulty 3/5

Car A heads east at 50 mph50~\text{mph}; car B heads north at 60 mph60~\text{mph}, both leaving the same intersection at t=0t = 0. After 11 hour, the distance between them grows at

  • A

    10 mph10~\text{mph}

  • B

    502+602 mph≈78.1 mph\sqrt{50^2 + 60^2}~\text{mph} \approx 78.1~\text{mph}

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  • C

    3000 mph3000~\text{mph}

  • D

    110 mph110~\text{mph}

Explanation

D2=x2+y2D^2 = x^2 + y^2, DD′=xx′+yy′D D' = x x' + y y'. At t=1t = 1: x=50x = 50, y=60y = 60, D=6100D = \sqrt{6100}. D′=(50⋅50+60⋅60)/D=6100/6100=6100≈78.1D' = (50 \cdot 50 + 60 \cdot 60)/D = 6100/\sqrt{6100} = \sqrt{6100} \approx 78.1.

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