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?

bigguy1989 May 28, 2018

#1**0 **

\(\boxed{\text{Note: } \\\text{speed} = \dfrac{\text{distance}}{\text{time}}\\\text{distance} = \text{speed}\cdot\text{time}\\\text{time}=\dfrac{\text{distance}}{\text{speed}}}\)

\(\text{Let }d\text{ miles be the distance between Washington DC and Baltimore.}\\\quad\;\;\;\!\;\! t\text{ hours be the time of return trip.}\)

(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 = \(d(\dfrac{1}{60}+\dfrac{1}{50})\)

\(\quad\text{Average speed} \\= \dfrac{d}{d\left(\dfrac{1}{60}+\dfrac{1}{50}\right)}\\=\dfrac{1}{\dfrac{1}{60}+\dfrac{1}{50}}\\=\dfrac{300}{11}\text{mph}\)

.MaxWong May 28, 2018

#4**0 **

It printed it twice!!!

Guest May 28, 2018

edited by
Guest
May 28, 2018

edited by Guest May 28, 2018

edited by Guest May 28, 2018

edited by Guest May 28, 2018

edited by Guest May 28, 2018

#4**0 **

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.**

Guest May 28, 2018