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

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

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