2070.84 – Particle Motion

A particle moves in a straight line so that its speed is constant for a mile, then changes abruptly for the next mile, is constant for that mile, then changes again---and so on. The speed varies so that for each mile it's inversely proportional to the total integral miles previously traveled.

If the second mile is traversed in 2 hours, how long does the particle take to travel the nnth mile, for n=2,3,4,…n = 2, 3, 4, \ldots?

Solution

Make a chart along these lines:

 Mile  Prev. Miles  Rate 10r121r2=k132r3=k2⋮⋮⋮\begin{array}{c|c|c}\text{ Mile } & \text{ Prev. Miles } & \text{ Rate } \\\hline 1 & 0 & r_1 \\[0.5em]2 & 1 & r_2 = \dfrac{k}{1} \\[1em]3 & 2 & r_3 = \dfrac{k}{2} \\[0.5em]\vdots & \vdots & \vdots\end{array}

Here kk is a constant of proportionality. We are given that the first mile took two hours to traverse. So in fact,

r2=12=k1,r_2 = \dfrac{1}{2} = \dfrac{k}{1},

thus k=1/2k = 1/2. With this value established, we can complete the table:

 Mile  Prev. Miles  Rate 10r121r2=1232r3=1/22=1443r4=1/23=16⋮⋮⋮nn−1rn=1/2n−1=12(n−1)\begin{array}{c|c|c}\text{ Mile } & \text{ Prev. Miles } & \text{ Rate } \\\hline 1 & 0 & r_1 \\[0.5em]2 & 1 & r_2 = \dfrac{1}{2} \\[1em]3 & 2 & r_3 = \dfrac{1/2}{2} = \dfrac{1}{4} \\[1em]4 & 3 & r_4 = \dfrac{1/2}{3} = \dfrac{1}{6} \\[1em]\vdots & \vdots & \vdots \\[1em]n & n-1 & r_n = \dfrac{1/2}{n-1} = \dfrac{1}{2 (n-1)}\end{array}

For that nthth mile, d=1=rt=12(n−1)td =1 = rt = \dfrac{1}{2 (n-1)} t, so t=2(n−1). t = 2 (n-1).