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March 6th, 2017, 05:06 PM  #1 
Newbie Joined: Mar 2017 From: Tennessee Posts: 2 Thanks: 0  Speed question
How fast is one vehicle passing another if both are going the same direction but at different speeds. Using an example of one going 80 and the other going 40, it seems the quick easy answer would be the 80 vehicle is passing the other at 40. But if you think of it in terms of me going 40 and I pass a stationary object, then yes, I'm passing it at 40, I zoom by it quickly. But if I'm doing 80 and I pass you going the same direction doing 40, it no longer seems I zoom past you, though the difference in speed is still 40. Is this a simple perception issue or is there some equation to figure this. Another example is in the past I have been doing 50 and there is one particular car, a Honda civic, that on some models their speedometer is so large, I can see their speed clearly. Let's say they are doing 40. The difference in speed is 10 but it seems I am passing them at a much slower rate than if I were passing them when they are sitting still and I'm doing 10. Make sense? Is this perception or because they are moving the same direction am I crawling past them at 3 MPH? Ok, I'm starting to confuse even myself, I hope some one can set me straight, thanks for any help.

March 6th, 2017, 06:22 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs 
Let's use a more convenient system of measures such as feet and seconds. Suppose you are traveling at a constant rate of $v\,\frac{\text{ft}}{\text{s}}$ and your beloved grandmother is overtaking and passing you in her twin turbo Lamborghini Aventador traveling at a constant rate of $(v+a)\,\frac{\text{ft}}{\text{s}}$ where $0<a$...how long will it take for her to be $b$ ft ahead of you? During the time $t$ it takes for her to be $b$ ft in front of you, you will have traveled: $\displaystyle d=vt$ And meemaw will have traveled: $\displaystyle d+b=vt+b=(v+a)t\implies t=\frac{b}{a}$ We see then that $t$ does not depend on $v$ at all, only on $a$ and $b$...thus any difference is a matter of perception and not of reality. 
March 6th, 2017, 06:58 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,153 Thanks: 252 
Perception issue. The relative speeds are all that matters. Unless it's light, then it's all weird. If you are going forward and a beam of light passes you from behind, you still measure it at c relative to your speed ... no matter what speed you are going. Now how can that possibly be? If someone can explain that to me I'd be grateful. I can understand that once you make that assumption, you get the Fitzgerald contraction and all the rest of it, but why does light work that way? How does light, of all the stuff in the universe, know to behave that way? 
March 6th, 2017, 08:26 PM  #4 
Newbie Joined: Mar 2017 From: Tennessee Posts: 2 Thanks: 0 
Thanks for the quick replies 
March 6th, 2017, 08:33 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,773 Thanks: 607  

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