Two objects move along the same straight line. When they…

Two objects move along the same straight line. When they move away from each other, the distance between them changes by $\Delta d_1 = 16\ \mathrm{m}$ in $\Delta t_1 = 10\ \mathrm{s}$. When they move toward each other, the distance changes by $\Delta d_2 = 3\ \mathrm{m}$ in $\Delta t_2 = 3\ \mathrm{s}$. What are the speeds of the objects?

Step 1: Interpret the two situations

If two objects move in opposite directions, the distance between them changes at $v_1 + v_2$, whether they approach or move apart. The two rates here are different (1.6 m/s and 1 m/s), so the two situations cannot both be opposite-direction motion. The consistent reading is:

• moving apart in opposite directions: the gap changes at $v_1 + v_2$;
• moving in the same direction, the faster one approaching the slower: the gap changes at $v_1 - v_2$.

Step 2: The two equations

\[ v_1 + v_2 = \frac{\Delta d_1}{\Delta t_1} = \frac{16\ \mathrm{m}}{10\ \mathrm{s}} = 1.6\ \mathrm{m/s}, \]

\[ v_1 - v_2 = \frac{\Delta d_2}{\Delta t_2} = \frac{3\ \mathrm{m}}{3\ \mathrm{s}} = 1\ \mathrm{m/s}. \]

Step 3: Solve

Add the equations, then subtract them: $2v_1 = 2.6\ \mathrm{m/s} \;\Longrightarrow\; v_1 = 1.3\ \mathrm{m/s}$.

\[ 2v_2 = 0.6\ \mathrm{m/s} \;\Longrightarrow\; v_2 = 0.3\ \mathrm{m/s}. \]

\[v_1 = 1.3\ \mathrm{m/s}, \qquad v_2 = 0.3\ \mathrm{m/s}\]

Check

$ (1.3 + 0.3)\ \mathrm{m/s} \cdot 10\ \mathrm{s} = 16\ \mathrm{m}\ \checkmark $

$ (1.3 - 0.3)\ \mathrm{m/s} \cdot 3\ \mathrm{s} = 3\ \mathrm{m}\ \checkmark $