Average velocity of jumping into car

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Homework Help Overview

The discussion revolves around calculating the average speed required for a car after jogging a certain distance, with a specific overall average speed in mind. The problem is situated within the context of kinematics, specifically focusing on average velocity and time calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods to calculate average speed, questioning the validity of using the arithmetic mean versus the definition of average speed based on total distance and total time. There is a discussion about the calculations involved in determining the speed of the car after jogging.

Discussion Status

Some participants have provided calculations and reasoning, while others have raised questions about the correctness of these methods. There is an ongoing exploration of the definitions and relationships between average speed and the components of the journey.

Contextual Notes

Participants mention the use of a calculator for computations, with some expressing skepticism about its reliability. There are also references to homework constraints and the nature of the problem as part of a physics assignment.

tony873004
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[Walker2 2.P.016.] You jog at 6.2 mi/h for 5.0 mi, then you jump into a car and drive for another 5.0 mi. With what average speed must you drive if your average speed for the entire 10.0 miles is to be 11.0 mi/h?

I got 15.8, but the computer program (webassign) says that's wrong.

I did (6.2+x) / 2 = 11

6.2+x = 22
x=22 - 6.2
x = 15.8 mi/h

What did I do wrong?
 
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I jog for (5 mi)/(6.2 mi/h) and I drive for (5/v) hrs where v is the speed of driving, i.e of the car.

Hence average speed = total distance travelled/total time taken
= 10 mi/(0.806 hrs + (5/v) hrs) = 11.0 mi/h

So, 10/11 = 0.806 + 5/v = 0.909
which gives 5/v = 0.103 or v = 48.54 mi/h (the calculations were performed using windows calc so please check them).

Now you have used [tex]v_{avg} = \frac{v_{jog}+v_{car}}{2}[/tex].

By definition, [tex]v_{avg} = \frac{S_{total}}{T_{total}}[/tex] where S_total = total distance traveled and T = total time taken to travel this distance. Do you see a difference between the two? Do you think the two are equivalent here?

If you still have a doubt, write [tex]v_{jog} = S_{jog}T_{jog}[/tex] and [tex]v_{car} = S_{car}T_{car}[/tex]. But please check the reasoning, the definition and remind yourself that it is the average speed you're concerned about. If you have a further doubt, you might want to draw a rough graph of the entire motion (or a series of graphs of displacement vs time, velocity vs time).

Hope that helps,

Cheers
Vivek
 
Thank you, Maverick. Not only was your answer correct, but your explanation was helpful too.

Just curious, what's wrong with the Windows calculator that you'd have to tell me to check the numbers since you used it?
 
Hee hee...well its windoze after all from microwsoft and you never know if the answer is monopolized :smile:

Seriously though, I wanted you to check if my calculations are correct.

Cheers
Vivek
 
maverick280857 said:
Hee hee...well its windoze after all from microwsoft and you never know if the answer is monopolized :smile:

As a side note, our homework in my Visual C# class was to write a clone of the Windows calculator. I've got my own ideas as to how a calculator should be designed. The one I made has 5 memories which is very useful for these types of physics problems that often have several steps, and several values that will be needed again later in the problem.

Here's some screenshots:
http://orbitsimulator.com/csc202/calc1.GIF
http://orbitsimulator.com/csc202/calc2.GIF

and a downloadable .exe file for anyone who wants a copy of this calculator:
http://orbitsimulator.com/csc202/myCalculator.exe

I'm not sure if it will work on computers that don't have the .net framework installed though, since it was written in Microsoft Visual C#.
 
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