\section*{Task 2 - V11122}

The Octavia Touring Sport Car from Škoda Automobile Works Prague reaches a speed of \(80 \frac{\mathrm{km}}{\mathrm{h}}\) in 14 seconds after starting.

a) How many kilometers has it covered in this time (assuming uniform acceleration)? b) In what time, starting from the moment of the start, has it covered \(1 \mathrm{~km}\)? (Assume that the car continues to drive at the speed of \(80 \frac{\mathrm{km}}{\mathrm{h}}\) after reaching this speed.)
}

a) First, we look for the constant acceleration

\[
a(t)=a_{0}
\]

for all \(0 \mathrm{~s} \leq t \leq 14\) s. The velocity is given by

\[
v(t)=a_{0} t
\]

for all \(0 \mathrm{~s} \leq t \leq 14 \mathrm{~s}\) due to \(v(0 \mathrm{~s})=0 \frac{\mathrm{m}}{\mathrm{s}}\). Since

\[
v(14 \mathrm{~s})=a_{0} \cdot(14 \mathrm{~s})=80 \frac{\mathrm{km}}{\mathrm{h}}=\frac{80000}{3600} \frac{\mathrm{m}}{\mathrm{s}}=\frac{200}{9} \frac{\mathrm{m}}{\mathrm{s}}
\]

it follows that

\[
a_{0}=\left(\frac{200}{9} \frac{\mathrm{m}}{\mathrm{s}}\right) \cdot \frac{1}{14 \mathrm{~s}}=\frac{100}{63} \frac{\mathrm{m}}{\mathrm{s}^{2}}
\]

and thus for the distance traveled after 14 seconds

\[
w(14 \mathrm{~s})=\frac{a_{0}}{2} \cdot(14 \mathrm{~s})^{2}=\frac{19600}{126} \mathrm{~m}=\frac{1400}{9} \mathrm{~m} \approx 0.156 \mathrm{~km}
\]

Therefore, the vehicle covers a distance of approximately \(0.156 \mathrm{~km}\) during the acceleration phase.

b) For the total distance of \(1 \mathrm{~km}\), we have

\[
w_{\text {total }}=\frac{1400}{9} \mathrm{~m}+\left(\frac{200}{9} \frac{\mathrm{m}}{\mathrm{s}}\right) \cdot t_{2}=\frac{9000}{9} \mathrm{~m}
\]

Thus,

\[
\left(\frac{200}{9} \frac{\mathrm{m}}{\mathrm{s}}\right) \cdot t_{2}=\frac{7600}{9} \mathrm{~m}
\]

and therefore \(t_{2}=38 \mathrm{~s}\). Hence, \(1 \mathrm{~km}\) is covered after

\[
t_{\text {total }}=14 \mathrm{~s}+38 \mathrm{~s}=52 \mathrm{~s}
\]