Kinematics: the race that Lance could now win

In summary, Lance has been struggling to solve homework equations and is now 3 km behind Richard. Lance has tried to speed up but is still 3 km behind. Richard has completed the same distance in less time and so has already reached the finish line.
  • #1
quantum_enhan
23
0
yepp-1.png



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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  • #2
nobody?
 
  • #3
Here's what we know:


Lance has 6 km to go; [itex]D_L = 6[/tex]
Richard has some unknown distance remaining; [itex]D_R[/tex]
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])
Therefore, [itex]4V_{L1} = 6 - D_R[/tex]
But instead, Lance speeds up to 2 km/hr faster than Richard ([itex]V_{L2} = V_R + 2[/tex])
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])
 
  • #4
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] not [itex]6 - D_R = 4 V_{L1}[/tex]
 
  • #5
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]

2 min = [itex]\frac{1}{30}hr[/tex]


So,

[tex]D_L = 6[/tex]

[tex]\frac{V_R}{15} = 6 - D_R[/tex]

[tex]V_L = V_R + 2[/tex]

[tex]V_Lt = 6[/tex]

[tex]V_R(t - \frac{1}{30}) = D_R[/tex]
 
Last edited:
  • #6
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] not [itex]6 - D_R = 4 V_{L1}[/tex]

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]

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.

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.
 

Related to Kinematics: the race that Lance could now win

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the causes of that motion.

2. How does kinematics relate to Lance's race?

Kinematics is essential in understanding the speed, acceleration, and displacement of Lance during the race. It helps us analyze his performance and determine if he is on track to win.

3. What are the key factors that affect Lance's kinematics in the race?

There are several key factors that affect Lance's kinematics, including his initial velocity, the slope of the race course, air resistance, and his body's position and orientation during the race.

4. How can kinematics be used to predict Lance's race outcome?

By using kinematic equations and data from Lance's previous races, we can predict his future performance and determine if he has a chance of winning the race. We can also use kinematics to analyze his race strategy and make adjustments for optimal performance.

5. Can kinematics be applied to other sports besides cycling?

Yes, kinematics can be applied to any sport or physical activity that involves motion, such as running, swimming, or even throwing a ball. It is a fundamental concept in physics that applies to all types of motion.

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