Mary Hammonds

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?

abonirali59

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) $=\frac{\text{distance(d)}}{\text{speed (v)}}$ and the fact that the difference between the time of the two trips is 6 min, we can set the equation.

$T=\frac{150}{x+10}-\frac{120}{x}=\frac{1}{10}$
$T=\frac{x}{x}×\frac{150}{x+10}-\frac{x+10}{x+10}×\frac{120}{x}=\frac{1}{10}$
$T=\frac{150x}{x\left(x+10\right)}-\frac{120\left(x+10\right)}{x\left(x+10\right)}=\frac{1}{10}$
$T=\frac{150x-120\left(x+10\right)}{x\left(x+10\right)}=\frac{1}{10}$
$T=10\left(150x-120\left(x+10\right)\right)=x\left(x+10\right)$
$T={x}^{2}-290x=-12000$
So the quadratic equation is $T={x}^{2}-290x+12000=0$
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.
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}=\frac{-\left(-290\right)±\sqrt{{\left(-290\right)}^{2}-4\left(1\right)\left(12000\right)}}{2\left(1\right)}$
$x=\frac{290±\sqrt{84100-48000}}{2}=\frac{290±\sqrt{36100}}{2}=\frac{290±\sqrt{190×190}}{2}$
$x=\frac{290±190}{2}$
$x=\frac{290+190}{2}=\frac{480}{2}=240$ and $x=\frac{290-190}{2}=\frac{100}{2}=50$
So the speed of the first trip is can be either 50mi/h or 240 mi/h.

Ana Robertson