P01R2015 – Furiously Fast

We start by setting up the kinematic equations for each car, noting that these equations hold up to the moment that the cars use their charge (since their acceleration changes to $a_2$ at that instant).

Both cars starts and rest and at $x=0$ which means that we have $x_{01}=x_{02}=0$ and $v_{01}=v_{02}=0$. In addition, they have the same acceleration $a_0$. Thus, the kinematic equations for each car are

Car\ 1:\ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{x_1\left(t\right)=\frac{1}{2}a_0t^2+v_{01}t+x_{01}=\frac{1}{2}a_0t^2} \\
\\
v_1\left(t\right)=a_0t+v_{01}=a_0t \end{array}
\right.

Car\ 2:\ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{x_2\left(t\right)=\frac{1}{2}a_0t^2+v_{02}t+x_{02}=\frac{1}{2}a_0t^2} \\
\\
v_2\left(t\right)=a_0t+v_{02}=a_0t \end{array}
\right.

Car 1 uses its NOx charge at time $t_1=3.5\ s$ and its speed is then equal to

\boxed{v_1\left(t_1\right)=a_0t_1=10\cdot 3.5=35\ m/s}

Car 2 uses its NOx charge at time $t_2=6\ s$ and its speed is then equal to

\boxed{v_2\left(t\right)=a_0t_2=10\cdot 6=60\ m/s}

The position of car 1 at time $t_1=3.5\ s$ is equal to

\boxed{x_1\left(t_1\right)=\frac{1}{2}a_0t^2_1=\frac{1}{2}\cdot 10\cdot {3.5}^2=61.25\ m}

The position of car 2 at time $t_2=6\ s$ is equal to

\boxed{x_2\left(t_2\right)=\frac{1}{2}a_0t^2_2=\frac{1}{2}\cdot 10\cdot 6^2=180\ m}

For convenience, we summarize these results in the figure below

To determine how far ahead car 1 is when its charge runs out, we need to find the position of each car when this happens, keeping in mind that when the acceleration changes from $a_0$ to $a_{NOx}$ the kinematic equations need to be rederived to account for the new acceleration.

Car 1 engages its NOx charge at time $t_1=3.5\ s$ when its position is $x_{1N}=61.25\ m$ and its speed is $v_{1N}=35\ m/s$. The values serve as the initial position and initial velocity for the updated kinematic equations that describe the motion of car 1 and we write

Car\ 1:\ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{x_1\left(t\right)=\frac{1}{2}a_{NOx}t^2+v_{1N}t+x_{1N}} \\
\\
v_1\left(t\right)=a_{NOx}t+v_{1N} \end{array}
\right.

where the time $t$ is measured from the moment that car 1 engages its NOx charge (as opposed to the beginning of the race). The position of car 1 when its charge runs out is given by plugging into $\mathrm{\Delta }t=3\ s$ into $x_1\left(t\right)$ which yields

x_1\left(\mathrm{\Delta }t\right)=\frac{1}{2}a_{NOx}{\left(\mathrm{\Delta }t\right)}^2+v_{1N}\mathrm{\Delta }t+x_{1N}=\frac{1}{2}\cdot 14\cdot 3^2+35\cdot 3+61.25=229.25\ m

Since the NOx charge of car 1 runs out $6.5\ s$ after the beginning of the race, we conclude that car 2 spends $0.5\ s$ at an acceleration $a_{NOx}$ because it engages its charge at $t=6\ s$. Thus, we repeat the process to derive the kinematic equations for car 2.

Car 1 engages its NOx charge at time $t_2=6\ s$ when its position is $x_{2N}=180\ m$ and its speed is $v_{2N}=60\ m/s$. The values serve as the initial position and initial velocity for the updated kinematic equations that describe the motion of car 2 and we write

Car\ 2:\ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{x_2\left(t\right)=\frac{1}{2}a_{NOx}t^2+v_{2N}t+x_{2N}} \\
\\
v_2\left(t\right)=a_{NOx}t+v_{2N} \end{array}
\right.

where the time $t$ is measured from the moment that car 2 engages its NOx charge (as opposed to the beginning of the race). The position of car 2 after spending $0.5\ s$ at an acceleration $a_{NOx}$ is equal to

x_2\left(0.5\right)=\frac{1}{2}a_{NOx}t^2+v_{2N}t+x_{2N}=\frac{1}{2}\cdot 14\cdot {0.5}^2+60\cdot 0.5+180=211.75\ m

Finally, when the NOx charge runs out car 1 is ahead by a distance $d$ equal to

\boxed{d=x_1\left(\mathrm{\Delta }t\right)-x_2\left(0.5\right)=229.25-211.75=17.5\ m}

First, we summarize the information derived previously in the figure below

Now, to determine who wins the race we need to find out the total time it takes each car to cover the remaining distance that separates them from the finish line

Car 1 will reach a final speed $v’_1$ equal to

v'_1=a_{NOx}t+v_{1N}=14\cdot 3+35=77\ m/s

and will travel the remaining distance $D-x’_1=423-229.25=193.75 \ m$ at that speed which will take an amount of time $t’_1$ equal to

t'_1=\frac{D-x'_1}{v'_1}=2.51\ s

Car 1 therefore crosses the finish line at time $t_{1f}$ equal to

\boxed{t_{1f}=t_1+\mathrm{\Delta }t+t'_1=3.5+3+2.51=9.01\ s}

In the case of car 2, we first need to determine its position and speed when its charge runs out and therefore we compute $x_2\left(\mathrm{\Delta }t\right)$ and $v_2\left(\mathrm{\Delta }t\right)\ $as follows

\begin{aligned}
x_2\left(\mathrm{\Delta }t\right)&=\frac{1}{2}a_{NOx}{\left(\mathrm{\Delta }t\right)}^2+v_{2N}\left(\mathrm{\Delta }t\right)+x_{2N}=\frac{1}{2}\cdot 14\cdot 3^2+60\cdot 3+180=423\ m \\
\\
v_2\left(\mathrm{\Delta }t\right)&=a_{NOx}\mathrm{\Delta }t+v_{2N}=14\cdot 3+60=102\ m/s
\end{aligned}

Thus, we note that car 2 crosses the finish line exactly when its NOx charge runs out and that it therefore wins the race because its time to reach the finish line is equal to

\boxed{t_{2f}=t_2+\mathrm{\Delta }t=6+3=9.00\ s}

which is ever so slightly less than the time it takes car 1 to reach the finish line.