Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection. At the same

Question

Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 26 ft from the intersection. If the speeds of the cars at that instant of time are 8 ft/sec and 12 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.)

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Doris 1 month 2021-08-16T02:31:08+00:00 1 Answers 0 views 0

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    2021-08-16T02:32:16+00:00

    Answer:

    The answer to the question is;

    The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

    Explanation:

    We note that the distance  traveled by each car after 4 seconds is

    Car A = 19 ft in the west direction.

    Car B = 26 ft in the north direction

    The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the  west being the x coordinate.

    Therefore, let the distance between the two cars be s

    we have

    s² = x² + y²

    = (19 ft)² + (26 ft)² = 1037 ft²

    s = \sqrt{1037 ft^2} = 32.202 ft.

    The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

    Since s² = x² + y² we have

    \frac{ds^{2} }{dt} = \frac{dx^{2} }{dt}  + \frac{dy^{2} }{dt}

    2s\frac{ds }{dt} = 2x\frac{dx}{dt}  + 2y\frac{dy }{dt} which gives

    s\frac{ds }{dt} = x\frac{dx}{dt}  + y\frac{dy }{dt}

    We note that the speeds of the cars were given as

    Car B moving north = 12 ft/sec, which is the y direction and

    Car A moving west = 8 ft/sec which is the x direction.

    Therefore

    \frac{dy }{dt} =  12 ft/sec and

    \frac{dx}{dt} = 8 ft/sec

    s\frac{ds }{dt} = x\frac{dx}{dt}  + y\frac{dy }{dt} becomes

    32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec}  + 26ft\times 12\frac{ft}{sec}  = 464 ft²/sec

    \frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.} = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

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