A salesman drives from Ajax to Barrington, a distance of 120 mi, at a

Mary Hammonds

Mary Hammonds

Answered question

2022-01-11

A salesman drives from Ajax to Barrington, a distance of 120 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 min more time than the first leg, how fast was he driving between Ajax and Barrington?

Answer & Explanation

abonirali59

abonirali59

Beginner2022-01-12Added 35 answers

Solving this question, we can construct a quadratic equation that the represents the situation.
We can start by letting x equal the speed throughout th journey from Barrington the Collins to equal x+10
Using the time formula where time (t) =distance(d)speed (v) and the fact that the difference between the time of the two trips is 6 min, we can set the equation.
T=t2t1=6 min=660 hour=110
T=150x+10120x=110
T=xx×150x+10x+10x+10×120x=110
T=150xx(x+10)120(x+10)x(x+10)=110
T=150x120(x+10)x(x+10)=110
T=10(150x120(x+10))=x(x+10)
T=x2290x=12000
So the quadratic equation is T=x2290x+12000=0
We can solve this equation by factoring, completing the square, or using the quadratic formula.
Looking at the equation, we first specify the values of a,b,c to solve the eqaution using the Quadratic Formula.
a=1
b=290
c=12000
Then, we plug these values into the formula.
x=b±b24ac2a=(290)±(290)24(1)(12000)2(1)
x=290±84100480002=290±361002=290±190×1902
x=290±1902
x=290+1902=4802=240 and x=2901902=1002=50
So the speed of the first trip is can be either 50mi/h or 240 mi/h.
Ana Robertson

Ana Robertson

Beginner2022-01-13Added 26 answers

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