Average velocity of jumping into car

In summary, the conversation is about finding the average speed needed to achieve a total average speed of 11.0 mi/h after jogging for 5.0 mi at 6.2 mi/h and driving for 5.0 mi. The correct calculation for this is (6.2+x) / 2 = 11, which results in a speed of 15.8 mi/h. The conversation also includes a discussion on the difference between average speed and total speed and the use of a calculator in solving the problem.
  • #1
tony873004
Science Advisor
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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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  • #2
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
 
  • #3
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?
 
  • #4
Hee hee...well its windoze after all from microwsoft and you never know if the answer is monopolized :rofl:

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

Cheers
Vivek
 
  • #5
maverick280857 said:
Hee hee...well its windoze after all from microwsoft and you never know if the answer is monopolized :rofl:

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#.
 
Last edited by a moderator:

1. What is the average velocity of a person jumping into a car?

The average velocity of a person jumping into a car can vary depending on various factors such as the height of the person, the distance between the person and the car, and the speed at which the person jumps. However, on average, it can range from 2-3 meters per second.

2. How is the average velocity of jumping into a car calculated?

The average velocity of jumping into a car can be calculated by dividing the total distance traveled by the person while jumping into the car by the total time taken. This can be represented by the formula Vavg = d/t, where Vavg is the average velocity, d is the distance traveled, and t is the time taken.

3. Is the average velocity of jumping into a car the same as the speed of the car?

No, the average velocity of jumping into a car and the speed of the car are two different measurements. The average velocity of jumping into a car only takes into account the distance traveled and the time taken, while the speed of the car also considers the direction of motion.

4. How does the weight of a person affect their average velocity when jumping into a car?

The weight of a person can affect their average velocity when jumping into a car. Generally, a heavier person may require more force and energy to jump into a car, leading to a slower average velocity. However, other factors such as height, distance, and speed also play a significant role in determining the average velocity.

5. Can the average velocity of jumping into a car be increased?

Yes, the average velocity of jumping into a car can be increased by increasing the speed and distance at which the person jumps. Additionally, reducing the time taken to jump into the car can also increase the average velocity. However, it is important to ensure safety and proper technique when attempting to increase the average velocity of jumping into a car.

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