SOLUTION: Greg drove at a constant speed in a rainstorm for 258 miles. He took a​ break, and the rain stopped He then drove 208 miles at a speed that was 9 miles per hour faster than

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Question 1102120: Greg drove at a constant speed in a rainstorm for 258 miles. He took a​ break, and the rain stopped He then drove 208 miles at a speed that was 9 miles per hour faster than his previous speed. If he drove for 10 ​hours, find the​ car's speed for each part of the trip. THANK YOU!!
Found 3 solutions by htmentor, greenestamps, josmiceli:
Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
Let s = the speed and t = the time for the first part of the trip
Then s + 9 is the speed for the second part of the trip
The total time is 10 h, so the time for the second leg is 10 - t
For the first leg of the trip s = 258/t -> t = 258/s [1]
For the second leg, s + 9 = 208/(10 - t) [2]
We have two equations and two unknowns
Substitute [1] into [2]:
s + 9 = 208/(10 - 258/s) = 208s/(10s - 258)
(s + 9)(10s - 258) = 208s
10s^2 - 258s + 90s - 2322 - 208s = 0
10s^2 - 376s - 2322 = 0
5s^2 - 188s - 1161 = 0
This gives s = -5.4 and 43. Only the positive solution is valid.
Thus the other speed is 43 + 9 = 52
Ans: 43 mph and 52 mph

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


The fastest way to solve this problem is by trial and error, hoping that the speeds are whole numbers.


So a good guess for the first part of the trip is 6 hours at 43 mph.

That would leave 4 hours for the rest of the trip, which was 208 miles; 208 miles in 4 hours means 52 mph; and indeed 52mph is 9 mph faster than 43 mph.

Algebraically...

Let x and x+9 be the two speeds. Then 258 miles at speed x plus 208 miles at speed x+9 makes a total of 10 hours:








The lower speed is 43 mph; the higher speed is 52 mph.

Note that in the algebraic solution, we had to factor the quadratic x^2-37x-258; to do that, we had to find two numbers whose product was 258.

But that's exactly what we did in the first place. So the algebraic solution didn't make the work any easier; it only made us do more work (a LOT more!) to get to the answer.
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = time in hrs for 258 mi trip
= time in hrs for 208 mi trip
Let = speed in mi/hr for 258 mi trip
= speed in mi/hr for 208 mi trip
---------------------------------------
(1)
(2)
----------------------------------
(2)
(2)
and
(1)
Plug this into (2)
(2)
(2)
(2)
(2)
(2)
(2)
Use quadratic formula










and

--------------------
He drove 35.16 mi/hr for 1st part and
44.16 mi/hr for 2nd part
------------------------
Get another opinion unless you know
what answer is. Check math too.

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