202201211321 Problem 1.1

Two particles move along the x-axis uniformly with speeds v_1=8\,\mathrm{m/s} and v_2=4\,\mathrm{m/s}. At the initial moment the first point was 21\,\mathrm{m} to the left of the origin and the second 7\,\mathrm{m} to the right of the origin. When will the first point catch up with the second? Where will this take place? Plot the graph of the motion.

Extracted from A. A. Pinsky. (1980). Problems in Physics.


Set-up.

Rename the two particles by a and b. The velocity of particle a is \mathbf{v}_a=+8\,\mathrm{(m\, s^{-1})}\enspace\hat{\mathbf{i}} and that of particle b is \mathbf{v}_b=+4\,\mathrm{(m\, s^{-1})}\enspace\hat{\mathbf{i}}. The particles at time t=0 are located on the x-axis with x-coordinates x_a=-21 and x_b=+7 respectively.


Roughwork.

When t=0:

when t=1:

when t=2:

when t=3:

when t=4:

when t=5:

when t=6:

when t=7:


Solution.

The positions x_a(t), x_b(t) of particle a, b can be expressed in a function of discrete time interval

t=\{ t_i\in\mathbb{Z^{+}}\textrm{ s.t. } t_{i+1}-t_{i}=t_{i}-t_{i-1}=1\},

i.e.,

\begin{aligned} x_a(t_{i+1}) & =x_a(t_{i})+8 \\ x_b(t_{i+1}) & =x_b(t_{i})+4 \\ \end{aligned}

or simply, in continuous time intervals,

\begin{aligned} x_a(t) & =-21+8t \\ x_b(t) & = 7+4t \\ \end{aligned}

Particle a will meet particle b when x_a(t)=x_b(t) at some time t', as

\begin{aligned} x_a(t') & = x_b(t') \\ -21+8t' & = 7+4t' \\ 4t' & = 28 \\ t' & = 7 \\ \end{aligned}

so the place of meeting is

x_a(7)=-21+8(7)=\boxed{35}=7+4(7)=x_b(7).