Question

A train approaching a station does two successive half-kilometers in 16 and 20 seconds respectively. Assuming the retardation to be uniform, find the further distance the train runs before stopping.

Answer 1

We're going to use the fact that $s_1 = s_2 = 0.5~\text{km} = s$ to simplify some calculations.

Given times: $t_1 = 16~\text{s}, t_2 = 20~\text{s}.$

Let's assume the initial position of the train as $s_0 = 0.$ Then, the kinematic equations for the given reference positions:

$s = v_0t_1 + \frac{a}{2}t_1^2; \\ 2s = v_0(t_1+t_2) + \frac{a}{2}(t_1+t_2)^2; \\ s_{stop} = v_0t_{stop} + \frac{a}{2}t_{stop}^2; \\ v_{stop} = 0 = v_0t_{stop} + at_{stop}.$

Please pay attention that here the auxiliary values $s_{stop}, t_{stop}$ are counted from the initial position and time.

By excluding the values from the above system of equations,

$v_0 = s\frac{t_2^2+2t_1t_2-t_1^2}{t_1t_2(t_1+t_2)}; \\ a = -2s\frac{t_2-t_1}{t_2(t_2^2+3t_1t_2+2t_1^2)}; \\ t_{stop} = -\frac{v_0}{a}; \\ s_{stop} = -\frac{v_0^2}{2a} = \frac{1}{4}s\frac{(t_2^2+3t_1t_2+2t_1^2)(t_2^2+2t_1t_2-t_1^2)^2}{t_1^2t_2(t_2-t_1)(t_1+t_2)^2}.$

One can see that a is negative, which indicates retardation.

The further distance (after the first two portions) the train passed will be given by

${\Delta}s = s_{stop} - 2s. \\$

By entering the numerical values into the formula, and canceling all the seconds units,

${\Delta}s = 0.5~\text{km}~*~(\frac{1}{4}\frac{(20^2+3*16*20+2*16^2)(20^2+2*16*20-16^2)^2}{16^220(20-16)(16+20)^2} - 2) \approx 4.42~\text{km}.$