Two objects (1 and 2) start at the same moment from the same…

Two objects (1 and 2) start at the same moment from the same point and move in mutually perpendicular directions with speeds $v_1 = 30\ \mathrm{km/h}$ and $v_2 = 40\ \mathrm{km/h}$. How does the distance between them change with time? What is their distance when the first object has travelled $s_1 = 90\ \mathrm{km}$?

Explanation of Method

Two objects moving apart at right angles.

Step 1: Positions after time $t$

Put the start at the origin, object 1 moving along one axis and object 2 along the other: $s_1 = v_1 t = 30t$.

\[ s_2 = v_2 t = 40t \qquad (\text{km, with } t \text{ in h}). \]

Step 2: Distance with Pythagoras

The two paths are perpendicular, so the paths and the line joining the objects form a right triangle: \[d(t) = \sqrt{(v_1 t)^2 + (v_2 t)^2} = t\sqrt{v_1^2 + v_2^2} = t\sqrt{900 + 1600}.\]

\[d(t) = 50\,t\ \ \mathrm{km} \quad (t \text{ in hours})\]

The distance grows uniformly, by 50 km every hour, as if the objects were moving apart along a line at 50 km/h.

$v_1 = 30\ \mathrm{km/h}$ $v_2$ $30$ $60$ $90\ \mathrm{km}$ $40$ $80$ $120$ $50$ $100$ $d = 150\ \mathrm{km}$ $t = 1\ \mathrm{h}:\ 30,\ 40,\ 50$ $t = 2\ \mathrm{h}:\ 60,\ 80,\ 100$ $t = 3\ \mathrm{h}:\ 90,\ 120,\ 150$ $\text{legs and distance (km)}$

object 1 (east)
object 2 (north)

Every hour the right triangle grows by the same 30–40–50 step, so the distance grows by 50 km each hour. Drawn to scale.

Step 3: When object 1 has gone 90 km

\[ t = \frac{s_1}{v_1} = \frac{90\ \mathrm{km}}{30\ \mathrm{km/h}} = 3\ \mathrm{h}, \]

\[ d = 50 \cdot 3. \]

\[d = 150\ \mathrm{km}\]

Check

After 3 h: $ s_1 = 90\ \mathrm{km} $ and $ s_2 = 120\ \mathrm{km} $, so $ d = \sqrt{90^2 + 120^2} = 150\ \mathrm{km}\ \checkmark $

Result

\[\boxed{\,d(t) = 50\,t\ \ \mathrm{km} \quad (t \text{ in hours})\,}\]

\[\boxed{\,d = 150\ \mathrm{km}\,}\]

Remark

Note. The number $\sqrt{v_1^2 + v_2^2} = 50\ \mathrm{km/h}$ is the speed of one object as seen from the other (their relative speed). The general case with any angle between the paths is Translational Kinematics 025.