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For my first problem session in classical mechanics I got stuck on this problem about locomotive velocity. Here is the description:

A locomotive leaves a train station with the magnitude of its displacement along a straight track given by $s=(1.4 m/s^2)t^2$. Exactly $10$ s later, another locomotive on a parallel track passes the station with constant velocity. What minimum velocity must the second locomotive have so that it just catches the first locomotive without passing it? (Hint: Plot displacement versus time for both locomotives assuming the station is at $s=0$.)

I then proceeded to plot the graph of the first locomotive, and then I pretty much got stuck. I know that since the second locomotive has a constant velocity, its equation would be something like: $$s_2=vt ,$$ since there is no acceleration. I am supposed to choose an arbitrary time at which these two should meet and then set these two equations equal? ($s$ and $s_2$) this would give me a value for $v$, but it seems completely arbitrary.

I would appreciate some help. Thanks in advance.

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1 Answer 1

Write the two displacements. $s_1=1.4t^2$ and $s_2=v(t-10)$ and set them equal to each other. Since you have a quadratic equation in $t$, you know the conditions you must have to have a solution. In other words make sure the discriminant is positive, that should give you the answer.

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can you please explain how you got (t-10) ? shouldn't it be t+10, since train number 2 is starting 10 seconds later? –  Virtuoso Sep 16 '11 at 3:01
    
@Virtuoso: The time the second train travels past the station is 10 seconds less than the time the first train has been traveling. It is the length of the interval from 10 to t. –  Ross Millikan Sep 16 '11 at 3:59
    
I set s1 and s2 equal to each other, I get 1.4t^2=v(t-10). Then I get v = 1.4t^2/(t-10). Plugging that back into s2 you get the same answer as s1. What am I doing wrong? –  Virtuoso Sep 18 '11 at 1:41

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