SOLUTION: Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up. If th
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up. If th
Log On
Question 610524: Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up. If the total trip took 1 hour, then what was Jose’s average speed on the return trip? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Jose drove 15 miles to pick up his sister and then returned home.
On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up.
If the total trip took 1 hour, then what was Jose’s average speed on the return trip?
:
Let s = speed on the return trip
then
(s-15) = speed on the initial trip
:
Write time equation: time = dist/speed
:
return time + initial time = 1 hr + = 1
multiply by s(s-15), results
15(s-15) + 15s = s(s-15)
15s - 225 + 15s = s^2 - 15s
30s - 225 = s^2 - 15s
:
combine on the right
0 = s^2 - 15s - 30s + 225
:
A quadratic equation
s^2 - 45s + 225 = 0
:
Use the quadratic formula to find s
In this equation x=s; a=1; b=-45; c=225
:
:
Two solutions. but only one will make sense
s =
s = 39.27 mph on the return trip
:
:
You can check this by finding the times + =
