Jenny dirves 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there. If her total journey time to get to work and back is 1 hour, what was her average speed on the way to work?
We have s, s-20, t and 1-t.
We calculate the time in both directions, based on the speed.
t = 24/s
1-t = 24/(s-20)
We know that the rouind trip is one hour.
t + (1-t) = 1
Substitute the equivalent values for time, using speed.
24/s + 24/(s-20) = 1
Multiply both sides of the equation by s * (s-20) to eliminate the fractions.
s(s-20) * (24/s + 24/(s-20)) = s(s-20) * 1
(s-20) * 24 + s * 24 = s(s-20)
24s - 480 + 24s = s^2 -20s
48s - 480 = s^2 - 20s
-480 = s^2 - 20s - 48s
-480 = s^2 - 68s
Re-arrange it so s is on the left.
s^2 - 68s = -480
Use "complete the square" to proceed.
s^2 - 68s + (68 * 1/2)^2 = -480 + (68 * 1/2)^2
s^2 - 68s + 34^2 = -480 + 1156
(s - 34) * (s - 34) = 676
Take the square root of both sides.
s - 34 = 26
s = 26 + 34
s = 60
Jenny's average speed on the way to work is 60mph.
We need to check the answer. 60 - 20 = 40mph going home.
24mi / 60mph = 0.4 hr
24mi / 40mph = 0.6 hr
Together, the two parts of the trip add to 1 hr.