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?

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