Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Mario is racing with Wario on Moo Moo Meadows when a goomba, ready to avenge all of his friends' deaths, came and hijacked Mario's kart. A graph representing the motion of Mario at any instant is shown below. The velocity acquired by Mario is shown on the x-axis, and the net power of his movement is shown on the $\mathrm{y}$-axis. When Mario's velocity is $6 \mathrm{~m} / \mathrm{s}$, he eats a mushroom which gives him a super boost.

<img_4272>

You may need to make measurements. Feel free to print this picture out as the diagram is drawn to scale. Find the total distance from Mario runs from when his velocity is $0 \mathrm{~m} / \mathrm{s}$ to when his velocity just reaches $9 \mathrm{~m} / \mathrm{s}$ given that Mario's mass is $m=89 \mathrm{~kg}$. Answer in meters and round to one significant digit.

Solution:
Our first goal is to find an expression for the power curve $P(v)$. To do this, let us select a few points on the curve. The easiest point to pick is $(0,0)$ since it is fixed at the origin. The next two easiest points to pick are those that are on the lines $x=3$ and $x=6$ which are given as approximately $(3,1.8)$ and $(6,7.2)$. Note that $y$-axis is in units of $100 \mathrm{~W}$ so in reality these two points are given as $(3,180)$ and $(6,720)$. This curve is resemblant of a quadratic in the form of $y=k_{1} x^{2}$ and upon solving for $k_{1}$ we find that the curve is given as $P=20 v^{2}$. Secondly, the next line remains constant with respect to time as a line $P=900$. Therefore, we can write a piecewise function for power defined by
$$
P(v)=\left\{\begin{array}{ll}
20 v^{2} & \text { if } v \geq 0, \quad \text { and } \quad v<6 \\
900 & \text { if } v \geq 6
\end{array} .\right.
$$



We need to find the relationship between power, velocity, and displacement of Mario. Consider dividing the displacement into tiny rectangular pieces with width $\Delta t$ such that

$$
s=\sum_{i \in \mathbb{N}} \Delta s_{i}=\sum_{i} v_{i}(t) \cdot \Delta t
$$

We want to the displacement to be expressed in terms without $\Delta t$. This means that we have to find a relationship for $\Delta t$. Note that

$$
\Delta t=\Delta v \cdot \frac{\Delta t}{\Delta v}=\Delta v \cdot \frac{1}{\Delta v / \Delta t}=\frac{\Delta v}{a}
$$

Therefore, we now know the displacement to be expressed as

$$
s=\sum_{i \rightarrow 0} v_{i}(t) \frac{\Delta v}{a}=\int \frac{v}{a(v)} d v
$$

To find an expression for $a(v)$ in terms of power, we note that

$$
P(v)=F(v) \cdot v \Longrightarrow a(v)=\frac{P(v)}{v m}
$$

which means that upon substituting,

$$
s=\int \frac{v^{2} m}{P(v)} d v \Longrightarrow s=\int_{0}^{6} \frac{89 v^{2}}{20 v^{2}} d v+\int_{6}^{9} \frac{89 v^{2}}{900} d v=43.61 \approx 40 \mathrm{~m}
$$

Final answer: 40 m