Kinematics: the race that Lance could now win

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Homework Help Overview

The discussion revolves around a kinematics problem involving two racers, Lance and Richard, where the goal is to determine the dynamics of their race given specific distances and time differences. The original poster attempts to analyze their speeds and the time it takes for each to reach the finish line.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the distances and speeds of Lance and Richard, questioning the interpretation of the time difference and the distances remaining for each racer. There are attempts to set up equations based on the information provided, including the conversion of time units for clarity.

Discussion Status

Several participants are actively engaging with the problem, offering corrections and clarifications regarding the assumptions made about the time advantage Richard has over Lance. There is an ongoing exploration of the correct kinematic equations to use, with some guidance provided on unit conversions and the implications of the time difference.

Contextual Notes

Participants note the importance of correctly interpreting the time advantage and the distances involved, as well as the necessity of converting units to maintain consistency in calculations. There is an acknowledgment of the complexity introduced by the differing speeds and the remaining distances for each racer.

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Homework Equations


v=d/t


The Attempt at a Solution



VL=VR+2km/h
d=6km
tL1tR+4min
tL2=tR +2

VL= 6km / (2-4) = 3km/min which is OBVIOUSLY WRONG. I have tried numerous things for the past couple hours, but can't seem to figure it out/. Help?
 
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Here's what we know:


Lance has 6 km to go; [itex]D_L = 6[/tex]<br /> Richard has some unknown distance remaining; [itex]D_R[/tex]<br /> At Lance's current speed ([itex]V_{L1}[/tex]), it will take him 4 min to reach Richard's current position ([itex]6 - D_R[/tex])<br /> Therefore, [itex]4V_{L1} = 6 - D_R[/tex]<br /> But instead, Lance speeds up to 2 km/hr faster than Richard ([itex]V_{L2} = V_R + 2[/tex])<br /> At this speed, Richard will reach the finish line 2 min. before Lance ([itex]V_{L2}t =6[/tex], and [itex]V_R(t-2) = D_R[/tex])[/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]
 
Correction?

I believe that the statement that Richard is 4 minutes ahead of Lance, is to mean that it took Richard 4 minutes to cover the distance currently between them.

Therefore [itex]6 - D_R = 4V_R[/tex] <b>not</b> [itex]6 - D_R = 4 V_{L1}[/tex][/itex][/itex]
 
Whoops! We need to get our units straight. It's probably best to convert distances to km and times to hours in order to keep numbers small

4 min = [itex]\frac{1}{15}hr[/tex]<br /> <br /> 2 min = [itex]\frac{1}{30}hr[/tex]<br /> <br /> <br /> So,<br /> <br /> [tex]D_L = 6[/tex]<br /> <br /> [tex]\frac{V_R}{15} = 6 - D_R[/tex]<br /> <br /> [tex]V_L = V_R + 2[/tex]<br /> <br /> [tex]V_Lt = 6[/tex]<br /> <br /> [tex]V_R(t - \frac{1}{30}) = D_R[/tex][/itex][/itex]
 
Last edited:
zgozvrm said:
Correction?

I believe that the statement that Richard is 4 minutes ahead of Lance, is to mean that it took Richard 4 minutes to cover the distance currently between them.

Therefore [itex]6 - D_R = 4V_R[/tex] <b>not</b> [itex]6 - D_R = 4 V_{L1}[/tex][/itex][/itex]
[itex][itex] <br /> Yes! If Richard is 4 minutes ahead of Lance, that means that 4 minutes ago, Richard was where Lance is now. So, during that 4 minutes, Richard covered a distance of [itex]V_R \times 4min[/tex]<br /> <br /> Think of an observer sitting on the side of the course. When Richard passes by, he starts his stopwatch. When Lance finally passes that same point, he finds that 4 minutes have elapsed.<br /> <br /> Understanding this is key to being able to answer the question (followed by the ability to correctly write out the kinematic equations using the data given). When solved, you'll find that you get nice, neat numbers.[/itex][/itex][/itex]
 

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