Kinematics: the race that Lance could now win

[quantum_enhan]

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?

 

[quantum_enhan]

nobody?

 

[zgozvrm]

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]

 

[zgozvrm]

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]

 

[zgozvrm]

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 /> $$D_L = 6$$<br /> <br /> $$\frac{V_R}{15} = 6 - D_R$$<br /> <br /> $$V_L = V_R + 2$$<br /> <br /> $$V_Lt = 6$$<br /> <br /> $$V_R(t - \frac{1}{30}) = D_R$$[/itex][/itex]

 

[zgozvrm]

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]