Answer:
60 mph
Step-by-step explanation:
time = distance/speed
If s represents the return speed, then the relationships of the travel times is ...
300/(s-10)-1 = 300/s
300s -s(s-10) = 300(s-10) . . . . . multiply by s(s-10)
-s^2 +310s = 300s -3000 . . . . eliminate parentheses
s^2 -10s -3000 = 0 . . . . . . . . . . write in standard form
(s -60)(s +50) = 0 . . . . . . . . . . . . factor
This has solutions s=60, s=-50. The negative solution is extraneous.
The return speed was 60 mph.
We want to find the speed in your return given that increasing the speed by 10 mph would saved you an hour of trip.
The speed was 72.62 mi/h
We know the relationship:
distance = speed*time
Let's assume the speed is represented with the variable S, and the time it took you to return is represented with T.
We know that the distance of the roundtrip was 600 miles.
then we can write:
600mi = S*T
Now we know that if we increase the speed by 10mi/h, the time decreases by one hour, so we can also write:
600mi = (S + 10mi/h)*(T - 1h)
So we have a system of equations:
To solve this, we can isolate one of the variables in one of the equations and then replace that on the other equation.
I will isolate T in the first one:
T = 600mi/S
Now we replace this in the other equation to get:
600mi = (S + 10mi/h)*(600mi/S - 1h)
Now we can solve this for S, the speed.
600mi = 600mi - 1h*S + (6,000 mi^2/h)/S - 10mi
0 = -1h*S + (6,000 mi^2/h)/S - 10mi
Now we multiply both sides by S to get:
0 = -1h*S^2 + (6,000 mi^2/h) - 10mi*S
This is a quadratic equation, we can solve this byt using the Bhaskara's formula:
[tex]S = \frac{10mi \pm \sqrt{(10mi)^2 - 4*(6,000 mi^2/h)*(-1h)} }{2*(-1h)} \\\\S = \frac{ 10mi \pm 155.24mi}{-2h}[/tex]
We need to take the positive solution, so we get:
S = (10mi - 155.24 mi)/(-2h) = 72.62 mi/h
The speed was 72.62 mi/h
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