Algebra 2 Word Problem--Distance/rate/time?
A family drove 1080 miles to a vacation lodge. One the way back home their average speed was 6 mph less and their trip took 2.5 hours longer than on the way to the lodge. What was their average speed on their way TO THE LODGE?
thanks:)
Comments
Recall about speed: s = d/t → where:
d = distance
t = time
A family drove 1080 miles to a vacation lodge.
s1 = d1/t1
s1 = 1080/t1
1080 = s1 * t1
One the way back home their average speed was 6 mph less and their trip took 2.5 hours longer.
s2 = d2/t2
s2 = 1080/t2
1080 = s2 * t2
You can write: 1080 = 1080
s1 * t1 = s2 * t2 → but you know that: s2 = s1 - 6
s1 * t1 = (s1 - 6) * t2 → but you know that: t2 = t1 + 2.5
s1 * t1 = (s1 - 6) * (t1 + 2.5)
s1.t1 = s1.t1 + 2.5s1 - 6t1 - 15
2.5s1 - 6t1 - 15 = 0 → but you know that: s1 = 1080/t1
(2.5 * 1080/t1) - 6t1 - 15 = 0
(2700/t1) - 6t1 - 15 = 0 → you multiply by t1
2700 - 6t1² - 15t1 = 0 → you multiply by - 1
6t1² + 15t1 - 2700 = 0 → you simplify by 6
t1² + 2.5t1 - 450 = 0
Polynomial like: ax² + bx + c, where:
a = 1
b = 2.5
c = - 450
Δ = b² - 4ac (discriminant)
Δ = 2.5² - 4(1 * - 450) = 6.25 + 1800 = 1806.25 = 42.5²
x1 = (- b - √Δ) / 2a = (- 2.5 - 42.5) / (2 * 1) → no possible, because t1 must be > 0
x2 = (- b + √Δ) / 2a = (- 2.5 + 42.5) / (2 * 1) = 40/2 = 20
→ t1 = 20 h
Recall: t2 = t1 + 2.5
→ t2 = 22.5 h
Recall: s1 = 1080/t1
→ s1 = 54 mph (this is the average speed on their way to the lodge)
Recall: s2 = s1 - 6
→ s2 = 48 mph (this is the average speed on their way back home)