## Task 4 - 090814

On two adjacent lanes, two 4 m long trucks were driving in the same direction. The first had a speed of $60 \frac{\mathrm{km}}{\mathrm{h}}$, the second a speed of $70 \frac{\mathrm{km}}{\mathrm{h}}$. The second truck passed the first.

At the beginning of the considered process, the rear end of the first truck was $a=20 \mathrm{~m}$ in front of the front end of the second (see figure a); at the end of the process, the front end of the first was $a=20 \mathrm{~m}$ behind the rear end of the second (see figure b).

How long did this process take, and what distance was covered by the front end of the second truck during this time?
a)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0630.jpg?height=377&width=255&top_left_y=1459&top_left_x=772)
b)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0630.jpg?height=379&width=254&top_left_y=1461&top_left_x=1095)
Since the process has taken place according to the task, a solution exists. Let the following designations be introduced:

Speed of $K_{1}$ and $K_{2}: v_{1}=60 \frac{\mathrm{km}}{\mathrm{h}}, v_{2}=70 \frac{\mathrm{km}}{\mathrm{h}}$

Distance traveled by the front edge of $K_{1}$ and $K_{2}$ during the process: $s_{1}, s_{2}$

Duration of the process: $t$

Distance between the front edges at the beginning and at the end of the process: $d_{A}=d_{E}=24 \mathrm{~m}$

Sum of these distances: $D=d_{A}+d_{E}=48 \mathrm{~m}$

Assuming these quantities are those occurring in the process described in the task, it follows that $s_{1}=v_{1} t, \quad s_{2}=v_{2} t, \quad s_{1}+D=s_{2}$. This further implies $v_{1} t+D=v_{2} t$, thus

$$
t=\frac{D}{v_{2}-v_{1}}=\frac{0.048}{10} \mathrm{~h}=17.28 \mathrm{~s} \approx 17.3 \mathrm{~s}
$$

and

$$
s_{2}=v_{2} t\left(=\frac{v_{2} D}{v_{2}-v_{1}}\right)=336 \mathrm{~m}
$$

Solutions of the 1st Round 1969 adopted from [5]

### 5.11.2 2nd Round 1969, Class 8