A salesman drives from Ajax to Barrington, a distance of 150 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 183 mi from Barrington to Collins. If the second leg of his trip took 3 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 1st leg = 150/x
time for 2nd leg = 183/(x+10)
183/(x+10) - 150/x = 3/60 = 1/20
times 20x(x+10)
3660x - 3000(x+10) = x(x+10)
3660x - 3000x - 30000 = x^2 + 10x
x^2 - 650x + 30000 = 0
(x-600)(x - 50) = 0
x = 600 or x = 50
reasonable answer: x = 50 mph for 1st leg
check:
time for 1st leg = 150/50 = 3 hrs
time for 2nd leg = 183/60 = 3.05 hrs
difference = .05 hrs or .05(60) = 3 minutes
2nd answer: x = 600
time for 1st = 150/600 = .25 hrs
time for 2nd = 186/610 = .3 hrs
difference = .3-.25 = .05 hrs = 3 minutes
Unless he was "driving " a jet, let's go with the 50 mph
To solve this problem, let's first find the time it takes for the salesman to drive from Ajax to Barrington.
Let's assume the initial speed of the salesman between Ajax and Barrington is "x" miles per hour. We can use the formula: time = distance / speed.
The distance between Ajax and Barrington is given as 150 miles, and the speed is "x" miles per hour. So, the time taken for the first leg of the trip is:
time₁ = 150 / x (eq. 1)
Now, the salesman increases his speed by 10 miles per hour for the second leg of the trip. The distance between Barrington and Collins is 183 miles. So, the time taken for the second leg of the trip is:
time₂ = 183 / (x + 10) (eq. 2)
According to the problem, the second leg of the trip took 3 minutes more than the first leg. We need to convert this 3 minutes into hours to match the units:
3 minutes = 3/60 hours = 1/20 hours
Therefore, we can set up the equation:
time₂ = time₁ + 1/20 (eq. 3)
Now, let's substitute the values from equations 1 and 2 into equation 3 to solve for "x":
183 / (x + 10) = 150 / x + 1/20
To simplify this equation, we can multiply all terms by 20x(x + 10) to eliminate the denominators:
20x(183) = (150)(x + 10) + x(x + 10)
Simplifying further:
3660x = 150x + 1500 + x² + 10x
Rearranging terms and simplifying:
x² + 22x - 3660 = 0
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula:
(x + 70)(x - 52) = 0
This gives us two possible values for "x": -70 and 52. Since the speed cannot be negative, we take the positive value:
x = 52
Therefore, the salesman was driving at a speed of 52 miles per hour between Ajax and Barrington.