Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobile that left

Question

Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobile that left city A was 10 km/hour less than the speed of the other automobile. If the first automobile had left city A 4 1/2 hours earlier than the other automobile left city B, then the two would have met 150 km away from B. Find the distance between A and B.

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Madeline 2 weeks 2021-10-13T04:06:27+00:00 2 Answers 0

Answers ( )

    0
    2021-10-13T04:07:48+00:00

    Answer:

    450km

    Step-by-step explanation:

    Take it that each automobile travels at 30 km an hour, for 150 km, meaning it will be 450 km apart.

    0
    2021-10-13T04:07:49+00:00

    Answer:

      450 km

    Step-by-step explanation:

    Equations

    We can define 3 variables: a, b, d. Let “a” and “b” represent the speeds of the cars leaving cities A and B, respectively. Let “d” represent the distance between the two cities. We can write three equations in these three variables:

    1. The relation between “a” and “b”:

      a = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B

    2. The relation between speed and distance when the cars leave at the same time:

      d = (a +b)·5 . . . . . . distance = speed × time

    3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is …

      distance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance

      4.5a + (150/b)(a +b) = d

    __

    Solution

    Substituting for d, we have …

      4.5a + 150/b(a +b) = 5(a +b)

      4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses

      5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side

    Now, we can substitute for “a” and solve for b.

      5b^2 + 0.5b(b-10) -150(b -10 +b) = 0

      5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses

      11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2

      (11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor

    The solutions to this equation are …

      b = 60/11 = 5 5/11 . . . and . . . b = 50

    Since b must be greater than 10, the first solution is extraneous, and the values of the variables are …

    • b = 50
    • a = b-10 = 40
    • d = 5(a+b) = 5(90) = 450

    The distance between A and B is 450 km.

    _____

    Check

    When the cars leave at the same time, their speed of closure is the sum of their speeds. They will cover 450 km in …

      (450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h

    __

    When car A leaves 4 1/2 hours early, it covers a distance of …

      (4.5 h)(40 km/h) = 180 km

    before car B leaves. The distance remaining to be covered is …

      450 km – 180 km = 270 km

    When car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in …

      (270 km)/(90 km/h) = 3 h

    In that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required.

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