A bus is driving along a straight highway at constant speed v. You have noticed the bus when it was at some point A. From what area near the highway can you catch up with this bus if your running speed is u < v? Draw this area for u = v/2. For problem 1.1.10^*

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.1.10^*.$ A bus is driving along a straight highway at constant speed $v$. You have noticed the bus when it was at some point $A$. From what area near the highway can you catch up with this bus if your running speed is $u < v$? Draw this area for $u = v/2$.

    <h3>Solution</h3>
    <p>
        
         <center>
        <figure>
          <img src="1.1.10.png" alt="1.1.10"
            loading="lazy" width="200" />
          <figcaption>
            Chasing a bus
          </figcaption>
        </figure>
        </center>

        <p>
          1. Let the bus is at point $A$, and the catching up person starts from point $B$ and runs perpendicular to the roadway $AC$. Let us introduce the notations: $AC = L, BC = h, AB = s.$
        </p>
           <center>
        <figure>
          <img src="sol.png" alt="1.1.10"
            loading="lazy" width="200" />
          <figcaption>
            The plane bounded by $\alpha$
          </figcaption>
        </figure>
        </center>
        
        <p>
          2. From right-angled triangle $ABC$ we have
        </p>
        <p style="text-align: center;">
         $$L = s \cdot cos \frac{\alpha}{2}\text{ и }h = s \cdot sin \frac{\alpha}{2}$$
        </p>
        <p>
          3. Travel time of bus $t_1$ and passenger $t_2$ before meeting at point $C$
        </p>
        <p style="text-align: center;">
         ${t}_{1}=\frac{{L}}{{v}}=\frac{{s}\cos(\alpha/2)}{{v}};\quad{t}_{2}=\frac{{h}}{{u}}=\frac{{s}\sin(\alpha/2)}{{u}}$
        </p>
        <p>
          where from
        </p>
        <p style="text-align: center;">
        $$\fbox{$\alpha = 2 \cdot \text{arcsin} \frac{u}{v}$}$$
        </p>

    </p>

    <h4>Answer</h4>
    <p>
        From the region bounded by the angle $α = 2 \\, \text{arcsin}(u/v)$ with vertex at the point $A$, bisected by the motorway
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
    </p>
</footer>