We introduce the surface mass density of the wheel $\rho$:
\[
\rho = \dfrac{m}{4a^2},
\]
then the moment of inertia of the wheel is calculated using the formula:
\[
I = \int r^2 \, dm = \int r^2 \rho \, dS = \iint r^2 \rho \, dx\,dy = \rho \iint (x^2 + y^2)\, dx\,dy,
\]
Let us consider the integral:
\[
\iint (x^2 + y^2)\, dx\,dy = \int_{-a}^{a} dx \int_{-a}^{a} (x^2 + y^2)\, dy.
\]

We evaluate the integrals with respect to $dx$ and $dy$ separately.

\[
\int_{-a}^{a} (x^2 + y^2)\, dy
= \left( x^2 y + \dfrac{y^3}{3} \right)\Big|_{-a}^{a}
= 2a x^2 + \dfrac{2a^3}{3},
\]

\[
\int_{-a}^{a} \left( 2a x^2 + \dfrac{2a^3}{3} \right) dx
= \left( 2a \dfrac{x^3}{3} + \dfrac{2a^3}{3} x \right)\Big|_{-a}^{a}
= \dfrac{8a^4}{3}.
\]

Thus,

Alternatively, one can use the formula for the moment of inertia of a rod. In this approach, the square can be considered as a system of rods, which leads to the same result as obtained previously.

Determine the vehicle's total kinetic energy $E_{\text{kin}}$ and the angular velocity $\omega_T$ of the wheels in the tops of the road bump. Express the answer in the terms of $a$, $m$, $M$ and $v_0$. 

The total kinetic energy of the vehicle is the sum of the translational kinetic energy and the rotational kinetic energy of the wheels (according to parallel axis theorem). The angular velocity of the wheels can be determined from the linear velocity of the vehicle and the distance from the wheel axis to the point of contact with the road:
\[
v(t) = \omega(t)\, r(t) \;\Rightarrow\; \omega(t) = \dfrac{v(t)}{r(t)}.
\]

When the car is at the top of the road bump, the distance from the axis to the point of contact is equal to $a$, therefore:

Then, to calculate the kinetic energy using parallel axis theorem:
\[
E_{\text{kin}} = E_{\text{translate}} + E_{\text{rotate}}
= \left( M + 2m \right)\dfrac{v_0^2}{2} + \dfrac{I \omega_T^2}{2},
\]
from which we obtain:

Using the energy conservation law, determine the angular velocity $\omega_{V}$ of the wheels in the valley. Express the answers in terms of $a$, $m$, $M$ and $v_0$.

The kinetic energy remains constant throughout the motion; therefore, the angular velocity of the wheels can be found from the distance between the axis and the point of contact. At the valley of the road, this distance is equal to $\sqrt{2}\,a$:
\[
E_{\text{kin}} = \left( M + 2m \right)\dfrac{v^2}{2} + \dfrac{I\omega^2}{2}
= \left( M + 2m \right)\dfrac{(\omega r)^2}{2} + \dfrac{I\omega^2}{2},
\]
\[
E_{\text{kin}} = \left( M + 2m \right)\dfrac{(\omega r)^2}{2} + \dfrac{2ma^2\omega^2}{2\cdot 3}
= \dfrac{\omega^2}{2}\left( r^2(M+2m) + a^2\dfrac{2m}{3} \right).
\]

Substituting $r = \sqrt{2}\,a$:
\[
E_{\text{kin}} = \omega_V^2 a^2 \dfrac{3M + 8m}{3},
\]
and equating this to the initial kinetic energy:
\[
\omega_V^2 a^2 \dfrac{3M + 8m}{3} = \dfrac{3M + 10m}{6}\, v_0^2,
\]
from which:

Since the point of contact between the wheel and the road is always located directly below the axle, we write the distance from the wheel axis to the level of the valley — it remains constant:
\[
GO = G' O' = G' T' + T' O'.
\]

Note that the angle of inclination of the road is equal to the angle of inclination of the tangent to the road; therefore,
\[
\tan \alpha = -y'(x).
\]

We differentiate the equation describing the shape of the road to compute the tangent of the inclination angle:
\[
y'(x) = -\dfrac{h}{2a}\left(e^{x/a} - e^{-x/a}\right),
\]
from which we obtain

We express $\cos \alpha$ in terms of $\tan \alpha$ using the fundamental trigonometric identity:
\[
\cos \alpha = \dfrac{1}{\sqrt{1+\tan^2 \alpha}}.
\]
Combining the results of parts B1 and B2, we obtain:
\[
\sqrt{2}a = a\sqrt{1+\tan^2\alpha} + k - h\cdot\dfrac{e^{x/a}+e^{-x/a}}{2},
\]
\[
\sqrt{2}a = a\sqrt{1+\left(\dfrac{h}{2a} (e^{x/a}-e^{-x/a})\right)^2} + k - h\cdot\dfrac{e^{x/a}+e^{-x/a}}{2},
\]
\[
\left(\sqrt{2}a - k + h\cdot \dfrac{e^{x/a}+e^{-x/a}}{2}\right)^2
= a^2 + \left(\dfrac{h}{2} (e^{x/a}-e^{-x/a})\right)^2.
\]

The equation must hold at every point of the road. Let us consider the point $x=0$, where $e^{x/a}=e^{-x/a}=1$:
\[
\left( \sqrt{2}a - k + h \right)^2 = a^2
\;\Rightarrow\;
\dfrac{k-h}{a} = \sqrt{2}-1.
\]

Then the equation takes the form:
\[
\left(a - h + h\cdot \dfrac{e^{x/a}+e^{-x/a}}{2}\right)^2
= a^2 + \left(\dfrac{h}{2} (e^{x/a}-e^{-x/a})\right)^2,
\]
\[
\left(a + h\left(\dfrac{e^{x/a}+e^{-x/a}}{2}-1\right)\right)^2
= a^2 + \left(\dfrac{h}{2} (e^{x/a}-e^{-x/a})\right)^2.
\]

Expanding the brackets:
\[
2ah\left(\dfrac{e^{x/a}+e^{-x/a}}{2}-1\right)
+ h^2 \left(\dfrac{e^{x/a}+e^{-x/a}}{2}-1\right)^2
= h^2 \left(\dfrac{e^{x/a}-e^{-x/a}}{2}\right)^2,
\]
\[
ah(e^{x/a}+e^{-x/a}-2) - h^2(e^{x/a}+e^{-x/a}) + h^2 = -h^2,
\]
\[
(ah - h^2)(e^{x/a}+e^{-x/a}) = 2ah - 2h^2,
\]
which is satisfied at any point when $a = h$, from which we obtain:
 

Knowing the values of $k$ and $h$, the equation describing the shape of the road takes the form:
\[
y(x) = a\left(\sqrt{2} - \dfrac{e^{x/a}+e^{-x/a}}{2} \right).
\]

To determine the length of the road bump, let us find the $x$-coordinates of the valleys, i.e., where $y=0$:
\[
0 = a\left(\sqrt{2} - \dfrac{e^{x/a}+e^{-x/a}}{2} \right)
\;\Rightarrow\;
2\sqrt{2} = e^{x/a}+e^{-x/a}.
\]

To solve this equation, introduce an auxiliary variable $z = e^{x/a}$:
\[
2\sqrt{2} = z + \dfrac{1}{z}
\;\Rightarrow\;
z^2 - 2\sqrt{2}\,z + 1 = 0,
\]
from which we obtain:
\[
z = \sqrt{2} \pm 1
\;\Rightarrow\;
e^{x/a} = \sqrt{2} \pm 1.
\]

Considering $x>0$, only one root remains:
\[
e^{x/a} = \sqrt{2}+1
\;\Rightarrow\;
\dfrac{x}{a} = \ln(\sqrt{2}+1).
\]

Thus, the coordinates of the valleys are:
\[
x_d = a\ln(\sqrt{2}+1),
\qquad
x_s = -a\ln(\sqrt{2}+1),
\]
from which we find the length of the road bump:

From part A3, we know that the angular velocity of the wheels can be expressed in terms of the distance from the axis to the point of contact with the road:
\[
\dfrac{\omega^2}{2} \left( r^2(M+2m) + a^2\dfrac{2m}{3} \right)
= \dfrac{3M + 10m}{6}\, v_0^2.
\]

The distance $r(x)$ can be expressed using the shape of the road:
\[
r(x) = \sqrt{2}a - y(x).
\]

Solving this system, we obtain: