SOLUTION: I really need help with this word problem. We're on Quadratic Equations so I'm assuming it relates. Here it is: Sonya drives 160 miles at a certain speed. After stopping at a

Question 79401: I really need help with this word problem. We're on Quadratic Equations so I'm assuming it relates. Here it is:
Sonya drives 160 miles at a certain speed. After stopping at a rest stop, she drives an additional 250 miles at a speed 15 mph slower than before the stop. If she drove 2 hours longer after the stop than before the stop, what was her speed before the stop? We need to round answer to the nearest tenth of a mile-per-hour.
Any help would be greatly appreciated. Thanks!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Sonya drives 160 miles at a certain speed. After stopping at a rest stop, she drives an additional 250 miles at a speed 15 mph slower than before the stop. If she drove 2 hours longer after the stop than before the stop, what was her speed before the stop? We need to round answer to the nearest tenth of a mile-per-hour.
:
Let s = speed before the rest stop
Time = 160/s
:
Then (s-15) speed after the rest stop
Time = 250/(s-15)
:
A time equation as given by the problem:
:
time before the r.stop + 2 hr = time after the r.stop
:
+ 2 =
:
Multiply equation by s(s-15) and you have:
160(s-15) + 2(s(s-15)) = 250s
:
160s - 2400 + 2s^2 - 30s = 250s
:
2s^2 + 160s - 30s - 250s - 2400 = 0
:
2s^2 - 120s - 2400 = 0
:
Simplify, divide by 2
s^2 - 60s - 1200 = 0
:
Will not factor, use the quadratic formula: a=1; b=-60; c=-1200
:
I assume you know how to do that:
I got s = -15.8 and s = 75.8, this is the solution we want.
:
Speed after the r.stop: 75.8 - 15 = 60.8 mph
:
:
We can check it, see if the times differ by 2 hrs.
160/75.8 = 2.11 hrs
250/60.8 = 4.11 hrs
:
Did this make sense to you? Any questions?

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