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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?

 May 28, 2018
 #1
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\(\boxed{\text{Note: } \\\text{speed} = \dfrac{\text{distance}}{\text{time}}\\\text{distance} = \text{speed}\cdot\text{time}\\\text{time}=\dfrac{\text{distance}}{\text{speed}}}\)

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

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

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

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 May 28, 2018
 #2
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i feel like you have to mulitple 300 by 2 before dividing by 11

 May 28, 2018
 #3
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That is right !!!.

Guest May 28, 2018
 #4
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It printed it twice!!!

 May 28, 2018
edited by Guest  May 28, 2018
edited by Guest  May 28, 2018
edited by Guest  May 28, 2018
 #4
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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

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