The distance between cities A and B is d = 200 km . At the…

The distance between cities $A$ and $B$ is $d = 200\ \mathrm{km}$. At the same moment, one car leaves city $A$ and another leaves city $B$, driving toward each other. The car from $A$ has a speed of $v_A = 60\ \mathrm{km/h}$, and the car from $B$ has a speed of $v_B = 40\ \mathrm{km/h}$. a) Draw a graph of the position of each car versus time. b) Using the graph, find the time and place at which the cars meet.

Explanation of Method

Two cars driving toward each other.

Step 1: Choose one reference point

To compare the two cars on one graph, measure every position $x$ from city $A$ toward city $B$. The car from $A$ starts at $x = 0$ and moves forward; the car from $B$ starts at $x = 200\ \mathrm{km}$ and moves back toward $A$:

\[ x_A(t) = v_A\,t = 60t, \qquad x_B(t) = d - v_B\,t = 200 - 40t \qquad (x \text{ in km},\ t \text{ in h}). \]

Step 2: (a) Position–time graph

Both graphs are straight lines: one rising (the car from $A$), one falling (the car from $B$). The car from $A$ reaches $B$ after $\tfrac{200}{60} \approx 3.3\ \mathrm{h}$; the car from $B$ reaches $A$ after $\tfrac{200}{40} = 5\ \mathrm{h}$.

$x\ (\mathrm{km\ from\ } A)$ $t\ (\mathrm{h})$ $0$ $1$ $2$ $3$ $4$ $5$ $40$ $80$ $160$ $120$ $200\,(B)$ $0\,(A)$ $M:\ t = 2\ \mathrm{h},\ x = 120\ \mathrm{km}$ $80\ \mathrm{km}$ $x_B = 200 - 40t$ $x_A = 60t$

car from A
car from B

Position of each car measured from city A. The car from A climbs from 0 toward 200 km; the car from B falls from 200 km toward 0. The lines cross where both cars are at the same place at the same time.

Step 3: (b) Meeting time and place

The cars meet where the lines cross, that is, when both are at the same position: \[v_A\,t = d - v_B\,t \;\Longrightarrow\; t = \frac{d}{v_A + v_B} = \frac{200\ \mathrm{km}}{60\ \mathrm{km/h} + 40\ \mathrm{km/h}} = 2\ \mathrm{h}.\]

\[ x = v_A\,t = 60\ \mathrm{km/h} \cdot 2\ \mathrm{h} = 120\ \mathrm{km}. \]

\[t = 2\ \mathrm{h}, \qquad 120\ \mathrm{km} \text{ from } A \ \ (80\ \mathrm{km} \text{ from } B)\]

Check

$ \text{car from } A: 60 \cdot 2 = 120\ \mathrm{km}, \qquad \text{car from } B: 40 \cdot 2 = 80\ \mathrm{km} $

$ 120\ \mathrm{km} + 80\ \mathrm{km} = 200\ \mathrm{km} = d\ \checkmark $

Result

\[\boxed{\,t = 2\ \mathrm{h}, \qquad 120\ \mathrm{km} \text{ from } A \ \ (80\ \mathrm{km} \text{ from } B)\,}\]

Remark

Why position and not distance travelled? Each car's distance travelled grows from zero, so those two graphs would both rise and never show the meeting. Measuring both positions from the same point ($A$) makes the meeting visible as the crossing of the lines. The formula $t = \dfrac{d}{v_A + v_B}$ is the same relative-velocity idea as in Translational Kinematics 006: the gap closes at $v_A + v_B$.