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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.1.28.$ On an icy section of the highway, the coefficient of friction between the wheels and the road is ten times less than on an un-icy one. How many times do you need to reduce the speed of the car so that the braking distance on an icy section of the highway remains the same?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/2/2.1.28/sol.png"
+loading="lazy" width="350" />
+<figcaption>
+Forces acting on the car
+</figcaption>
+</figure>
+</center>
+
+<p>
+1. The external force when the car is moving is the friction force $F = \mu mg$, therefore, without taking into account the resistance from the air, the dynamic equation of motion has the form
+$$ ma = \mu mg $$
+
+$$ a = \mu g $$
+2. The kinematic equations of motion in this case are represented as follows:
+$$ \left\{\begin{matrix}
+ v = v_0-at \\
+ x=v_0-\frac{at^2}{2}
+ \end{matrix}\right. $$
+
+$$ t=\frac{v_0}{a}=\frac{v_0}{\mu g} $$
+3. When substituting the values ​​of acceleration and time into the second equation, we obtain the equation for the braking distance of a car
+$$ x=\frac{v_0^2}{\mu g}-\frac{v_0^2}{2\mu g}=\frac{v_0^2}{2\mu g} $$
+
+$$ v_0 = \sqrt{2\mu gx} $$
+Hence the speed needs to be
+reduced by $\sqrt{10}$ times
+</p>
+ </p>
+
+ <h4>Answer: $\sqrt{10}$ times</h4>
+
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