Question 592150
drove 100 miles, then increased her speed by 25 miles per hour for 225 miles.
 If the second part of the drive took 1 hour longer than the first part, find her average speed.
:
Let s = speed the 1s 100 miles
then
(s+25) = speed on the last 225 miles
:
Write a time equation; time = dist/speed
{{{225/((s+25))}}} - {{{100/s}}} = 1
multiply by s(s+25)
s(s+25)*{{{225/((s+25))}}} - s(s+25)*{{{100/s}}} = s(s+25)
225s - 100(s+25) = s^2 + 25s
225s - 100s - 2500 = s^2 + 25s
125s - 2500 = s^2 + 25s
0 = s^2 + 25s - 125s + 2500
A quadratic equation
s^2 - 100s + 2500 = 0
Factors to:
(s-50)(s-50) = 0
s = 50 mph on the 1st 100 mi
then
75 mph on last 225 mi
:
Let a = average speed for the whole trip which was 325 mi
{{{100/50}}} + {{{225/75}}} = {{{325/a}}}
multiply by 150a to clear the denominators; results
3a(100) + 2a(225) = 150(325)
300a + 450a = 48750
750a = 48750
a = 65 mph is the average speed for the trip
:
:
Check this out with a time equation
100/50 + 225/75 = 325/65
2 + 3 = 5