A salesman drives from Ajax to Barrington, a distance of 130 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 160

Question

A salesman drives from Ajax to Barrington, a distance of 130 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 160 mi from Barrington to Collins. If the second leg of his trip took 4 min more time than the first leg, how fast was he driving between Ajax and Barrington?

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Alice 3 months 2022-02-10T07:46:24+00:00 1 Answer 0 views 0

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    2022-02-10T07:47:32+00:00

    Answer:

    50 mi/hr or 390 mi/hr

    Step-by-step explanation:

    Distance between Ajax and Barrington is 130mi.

    Distance between Barrington and Collins is 160mi

    The speed for the second leg is increased by 10mi/hr

    Time to travel the second leg is 4 mins more than the first leg.

    Let x be the speed through out the journey from Ajax to Barrington.

    The speed of the journey from Barrington to Collins = x + 10

    Recall that

    Speed = distance /time

    Time = Distance /speed

    The difference between the time of the two trips is 4 mins. Therefore

    T = t2 – t1 = 4 mins

    T = t2 – t1 = 4/60 = 1/15 hr

    T = 160/(x + 10) – 130/x = 1/15

    = [x(160) – 130(x+10)] / (x+10)x = 1/15

    = [160x – 130x – 1300] /(x+10)x= 1/15

    = (30x – 1300) / x^2 + 10x = 1/15

    = 15(30x – 1300) = x^2 + 10x

    T = 450x – 19500 = x^2 + 10x

    = x^2 + 10x – 450x + 19500 = 0

    T = x^2 – 440x + 19500 = 0

    Using quadratic formula

    a = 1

    b= -440

    c = 19500

    x = [-b +/- √b^2 -4ac]/2a

    x = [-(-440) +/- √(-440)^2 – 4(1)(19500)] /2(1)

    x = [440 +/- √193600 – 78000] / 2

    x = (440 +/- √115600) / 2

    x = (440 +/- 340) / 2

    x = (440 + 340) / 2 or (440 – 340)/2

    x = 780/2 or 100/2

    x = 390 or 50

    x = 50 mi/hr or 390mi/hr

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