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 Bar-
rington 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?
speed for first leg --- x mph
speed for 2nd leg = x+10 mph
time for first leg = 120/x
time for 2nd leg = 150/(x+10)
120/x - 150/(x+10) = 6/60 = 1/10
multiply each term by 10x(x+10)
10(x+10)(120) - 10x(150) = x(x+10)
1200x + 12000 - 1500x = x^2 + 10x
x^2 + 310x + 12000 = 0
I don't think it factors, use the quadratic formula to solve, reject the negative value of x
Let's assume that the salesman's initial speed from Ajax to Barrington was "x" miles per hour.
To find the time taken for the first leg, we can use the formula:
Time = Distance / Speed
For the first leg, the distance is 120 miles, and the speed is "x" miles per hour. Therefore, the time taken for the first leg is:
Time 1 = 120 miles / x miles per hour = 120/x hours
Now, let's find the speed for the second leg. The salesman increases his speed by 10 miles per hour, so the speed for the second leg is "x + 10" miles per hour.
Using the same formula, the time taken for the second leg is:
Time 2 = 150 miles / (x + 10) miles per hour = 150 / (x + 10) hours
According to the problem, the second leg took 6 minutes longer than the first leg. To convert minutes to hours, we divide by 60:
6 minutes / 60 = 1/10 hours
So, the equation becomes:
Time 2 = Time 1 + 1/10
Substituting the values of Time 1 and Time 2, we get:
150 / (x + 10) = 120 / x + 1/10
To solve this equation, we first need a common denominator:
150(x) = 120(x + 10) + 1/10(x)(x + 10)
Expanding and simplifying, we get:
150x = 120x + 1200 + x^2 + 10x
Combining like terms:
x^2 + 40x - 1200 = 0
Now, we can solve this quadratic equation:
Using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 1, b = 40, and c = -1200.
Calculating the discriminant (b^2 - 4ac):
b^2 - 4ac = 40^2 - 4(1)(-1200) = 1600 + 4800 = 6400
Taking the square root of the discriminant:
√(b^2 - 4ac) = √6400 = 80
Substituting the values into the quadratic formula:
x = (-40 ± 80) / (2*1)
x1 = (-40 + 80) / 2 = 40 / 2 = 20
x2 = (-40 - 80) / 2 = -120 / 2 = -60
Since the speed cannot be negative, we discard the negative value.
Therefore, the speed the salesman was driving between Ajax and Barrington was 20 miles per hour.
To solve this problem, let's break it down step by step:
Step 1: Find the time taken for the first leg.
Let's assume the speed of the salesman while driving from Ajax to Barrington is "x" mi/h.
Using the formula: Speed = Distance / Time, we can rearrange the formula to find the time taken.
Time taken for the first leg = Distance / Speed = 120 / x.
Step 2: Find the time taken for the second leg.
The salesman increased his speed by 10 mi/h for the second leg, so his speed is now (x + 10) mi/h.
Using the same formula, the time taken for the second leg is: 150 / (x + 10).
Step 3: Set up and solve the equation.
We are given that the second leg took 6 minutes more time than the first leg.
Converting 6 minutes to hours, we have 6 / 60 = 1 / 10 hour.
So, the equation is: Time taken for the second leg = Time taken for the first leg + 1 / 10.
150 / (x + 10) = 120 / x + 1 / 10.
Step 4: Solve for x.
To simplify the equation, let's get rid of the fractions by multiplying every term by 10x(x+10):
1500x = 1200(x+10) + x(x+10).
Expanding and simplifying the equation:
1500x = 1200x + 12000 + x^2 + 10x.
Rearranging and combining like terms:
x^2 + 290x - 12000 = 0.
This is a quadratic equation. We can solve it by factoring or using the quadratic formula.
Factoring the equation:
(x - 80)(x + 370) = 0.
So, x = 80 or x = -370.
Since speed cannot be negative, the salesman's speed from Ajax to Barrington was 80 mi/h.
Therefore, the salesman was driving at a speed of 80 mi/h between Ajax and Barrington.