One car departs from city A to city B every Delta t = 10 min…

One car departs from city $A$ to city $B$ every $\Delta t = 10\ \mathrm{min}$. The cars travel at a speed of $v = 60\ \mathrm{km/h}$. The distance between the cities is $d = 60\ \mathrm{km}$. a) Draw a graph of the distance travelled by a car versus time. b) At what speed should the fourth car in the column leave city $A$ so that it reaches city $B$ at the same time as the first car? When and where does this car overtake the cars ahead of it?

Step 1: Set up the motion of each car

Measure time $t$ from the moment the first car leaves $A$. Car number $n$ leaves $(n-1)$ intervals later:

\[ t_n = (n-1)\,\Delta t = \frac{n-1}{6}\ \mathrm{h}. \]

Each car needs

\[ t_{\text{travel}} = \frac{d}{v} = \frac{60\ \mathrm{km}}{60\ \mathrm{km/h}} = 1\ \mathrm{h}, \] and its distance from $A$ while it is driving is

\[ s_n(t) = v\,(t - t_n) = 60\left(t - \frac{n-1}{6}\right)\ \mathrm{km}, \qquad t_n \le t \le t_n + 1\ \mathrm{h}. \]

Step 2: (a) Distance–time graph

All cars have the same speed, so their graphs are parallel straight lines with slope $60\ \mathrm{km/h}$, each shifted 10 min to the right. On the road, neighbouring cars are always $v\,\Delta t = 60 \cdot \tfrac{1}{6} = 10\ \mathrm{km}$ apart.

Step 3: (b) Speed of the fourth car

The first car arrives at $B$ at $t = 1\ \mathrm{h}$. The fourth car leaves at $t_4 = \tfrac{3}{6}\ \mathrm{h} = 30\ \mathrm{min}$, so it has only half an hour for the whole 60 km:

\[ v_4 = \frac{d}{1\ \mathrm{h} - 0.5\ \mathrm{h}} = \frac{60\ \mathrm{km}}{0.5\ \mathrm{h}} = 120\ \mathrm{km/h}. \]

Step 4: (b) When and where it overtakes the others

The fast car's position is $s_4(t) = 120\,(t - 0.5)$. It meets car $n$ when their positions are equal: \[120\,(t - 0.5) = 60\left(t - \frac{n-1}{6}\right) \;\Longrightarrow\; 120t - 60 = 60t - 10(n-1) \;\Longrightarrow\; t = 1 - \frac{n-1}{6}\ \mathrm{h}.\]

For car 3 ($n = 3$): $t = 1 - \tfrac{2}{6} = \tfrac{2}{3}\ \mathrm{h} = 40\ \mathrm{min}$, and $s = 120\left(\tfrac{2}{3} - \tfrac{1}{2}\right) = 120 \cdot \tfrac{1}{6} = 20\ \mathrm{km}$. The same steps for cars 2 and 1 give:

\[\begin{aligned} &v_4 = 120\ \mathrm{km/h} \\ &\text{passes car 3 at } 40\ \mathrm{min},\ 20\ \mathrm{km} \text{ from } A \\ &\text{passes car 2 at } 50\ \mathrm{min},\ 40\ \mathrm{km} \text{ from } A \\ &\text{reaches } B \text{ with car 1 at } 60\ \mathrm{min} \end{aligned}\]