SOLUTION: A salesman drives from Ajax to Barrington, a distance of 140 mi, at a steady speed. He then increases his speed by 15 mi/h to drive the 180 mi from Barrington to Collins. If the se
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Question 939969: A salesman drives from Ajax to Barrington, a distance of 140 mi, at a steady speed. He then increases his speed by 15 mi/h to drive the 180 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?
You can put this solution on YOUR website! A salesman drives from Ajax to Barrington, a distance of 140 mi, at a steady speed.
He then increases his speed by 15 mi/h to drive the 180 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?
:
let s = his original speed
(s+15) = his increased speed
then time from Aj to Bar
and = time from Bar to Col
:
Bar/Col time - Aj/Bar time = 4 min - = (we have to convert 4 min to hrs) - = (reduce the fraction)
multiply equation by 15s(s+15), cancel the denominators and you have
15s(180) - 15(s+15)(140) = s(s+15)
2700s - (15s+225)(140) = s^2 + 15s
2700s - 2100s - 31500 = s^2 + 15s
600s - 31500 = s^2 + 15s
Combine on the right to form a quadratic equation
0 = s^2 + 15s - 600s + 31500
0 = s^2 - 585s + 31500
you can use the quadratic formula here a=1; b=-585; c=31500. however this will factor
(s-525)(s-60) = 0
s = 525, not reasonable
s = 60 mph, his speed from Aj to Bar
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:
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See if that checks out; (Bar/Col speed is 15 mph more or 75 mph)
180/75 = 2.40 hrs
140/60 = 2.3333 hrs
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differs by .067 hrs, convert to min; .067(60) ~ 4.0 min
