Answer
(a) Speed at first electric eye = $12.2\,\mathrm{m/s}$ and speed at the second electric eye =$16.4\,\mathrm{m/s}$
(b) Distance to first electric eye = 24.8\,\mathrm{m}
(c) Time to reach the first electric eye = $4.07\,\mathrm{s}$
Work Step by Step
(a) Considering the car's travel between the two electric eyes, we are given
$\Delta x=20.0\,\mathrm{m}$, $\Delta t=1.40\,\mathrm{s}$ and $a=3.00\,\mathrm{m/s^2}$.
The initial speed of the car for this data is simply its speed at the first electric eye.
Using the kinematic equation,
\begin{align*}
\Delta x&=v_0\Delta t+{1\over 2}a \Delta t^2\\
20.0&=v_0(1.40)+{1\over 2}(3.00)(1.40)^2\\
20.0&=1.40v_0+2.94\\
v_0&={17.1\over 1.40}\\
v_0&=12.2\,\mathrm{m/s}
\end{align*}
The speed of the car when it passes the first electric eye is $12.2\,\mathrm{m/s}$.
To find the speed of the car at the second electric eye can be found by computing the final speed for the above data,
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
&=(12.2)^2+2(3.00)(20.0)\\
&=149+120\\
&=269\\
v&=\sqrt{269}\\
v&=16.4\,\mathrm{m/s}
\end{align*}
(b) For the car's journey from the start to the first electric eye,
$v_0=0$, $a=3.00\,\mathrm{m/s^2}$, and $v=12.2\,\mathrm{m/s}$. Calculating the distance to the first electric eye:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
(12.2)^2&=0+2(3.00)(\Delta x)\\
149&=6.00\Delta x\\
\Delta x&={149\over 6.00}\\
\Delta x&=24.8\,\mathrm{m}
\end{align*}
So, the first electric eye is at a distance of $24.8\,\mathrm{m}$ from the start.
(c) To compute the time required to reach the first electric eye, we use the kinematic equation
\begin{align*}
v&=v_0+at\\
12.2&=0+(3.00)t\\
t&={12.2\over 3.00}\\
t&=4.07\,\mathrm{s}
\end{align*}
Therefore, it takes the car $4.07\,\mathrm{s}$ for the car to reach the first electric eye.
