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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2022-02-10T07:46:24+00:00
2022-02-10T07:46:24+00:00 1 Answer
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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