+99 For travel from Washington DC to Baltimore by car, a constant speed of 60 mph was maintained. For the return trip, there was a constant speed of 50 mph. What was the average speed over the whole two-way trip?
+9519 \(\boxed{\text{Note: } \\\text{speed} = \dfrac{\text{distance}}{\text{time}}\\\text{distance} = \text{speed}\cdot\text{time}\\\text{time}=\dfrac{\text{distance}}{\text{speed}}}\)
\(\text{Let }d\text{ miles be the distance between Washington DC and Baltimore.}\\\quad\;\;\;\!\;\! t\text{ hours be the time of return trip.}\)
(magic: we can solve that without knowing the actual distance and time)
\(t = \dfrac{d}{50}\)
Time taken by Washington DC -> Baltimore = d/60
Total time used = \(d(\dfrac{1}{60}+\dfrac{1}{50})\)
\(\quad\text{Average speed} \\= \dfrac{d}{d\left(\dfrac{1}{60}+\dfrac{1}{50}\right)}\\=\dfrac{1}{\dfrac{1}{60}+\dfrac{1}{50}}\\=\dfrac{300}{11}\text{mph}\)
It printed it twice!!!
Let the average speed = S
[1/(60) + 1/(50)]S =2 [2 being twice the distance covered]
Solve for S:
(11 S)/300 = 2
Multiply both sides of (11 S)/300 = 2 by 300/11:
(300×11 S)/(11×300) = 300/11×2
300/11×11/300 = (300×11)/(11×300):
(300×11)/(11×300) S = 300/11×2
300/11×2 = (300×2)/11:
(300×11 S)/(11×300) = (300×2)/11
(300×11 S)/(11×300) = (11×300)/(11×300)×S = S:
S = (300×2)/11
300×2 = 600:
S = 600/11- Average speed over two-way trip.
