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; D_L = 6[/tex]<br /> Richard has some unknown distance remaining; D_R[/tex]&lt;br /&gt; At Lance&amp;#039;s current speed (V_{L1}[/tex]), it will take him 4 min to reach Richard&amp;amp;#039;s current position (6 - D_R[/tex])&amp;amp;lt;br /&amp;amp;gt; Therefore, 4V_{L1} = 6 - D_R[/tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; But instead, Lance speeds up to 2 km/hr faster than Richard (V_{L2} = V_R + 2[/tex])&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; At this speed, Richard will reach the finish line 2 min. before Lance (V_{L2}t =6[/tex], and V_R(t-2) = D_R[/tex])

 

[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 6 - D_R = 4V_R[/tex] <b>not</b> 6 - D_R = 4 V_{L1}[/tex]

 

[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 = \frac{1}{15}hr[/tex]<br /> <br /> 2 min = \frac{1}{30}hr[/tex]&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; So,&lt;br /&gt; &lt;br /&gt; D_L = 6&lt;br /&gt; &lt;br /&gt; \frac{V_R}{15} = 6 - D_R&lt;br /&gt; &lt;br /&gt; V_L = V_R + 2&lt;br /&gt; &lt;br /&gt; V_Lt = 6&lt;br /&gt; &lt;br /&gt; V_R(t - \frac{1}{30}) = D_R

 

[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 6 - D_R = 4V_R[/tex] <b>not</b> 6 - D_R = 4 V_{L1}[/tex]

&lt;br /&gt; &lt;br /&gt; 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 V_R \times 4min[/tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; 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.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; 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&amp;amp;#039;ll find that you get nice, neat numbers.