Answer
(a) $t=5.00\,\mathrm{s}$
(b) $\Delta x=112\,\mathrm{m}$
(c) $a=4.50\,\mathrm{m/s^2}$
(d) $\Delta x=226\,\mathrm{m}$
Work Step by Step
(a) For the braking phase,
$v_0=45.0$, $v=0$ and $a=-9.00\,\mathrm{m/s^2}$
Using the kinematic equation:
\begin{align*}
v&=v_0+at\\
0&=45.0+(-9.00)t\\
t&={45.0\over 9.00}\\
t&=5.00\,\mathrm{s}
\end{align*}
Time taken during the braking phase is $5.00\,\mathrm{s}$.
(b) To calculate distance travelled during the braking phase, we use the kinematic equation:
\begin{align*}
\Delta x&=v_0t+{1\over 2}at^2\\
&=(45.0)(5.00)+{1\over 2} (-9.00)(5.00)^2\\
&=225-113\\
&=112\,\mathrm{m}
\end{align*}
(c) During the speeding-up phase,
$v_0=0$, $v=45.0$ and $t=15.0-5.00=10.0\,\mathrm{s}$
From the defintion of acceleration,
$$a={v-v_0\over\Delta t}={45.0-0\over 10.0}=4.50\,\mathrm{m/s^2}$$
The acceleration of the car during the speeding-up phase is $4.50\,\mathrm{m/s^2}$
(d) To compute the distance covered during the speedin-up phase, we use the kinematic equation:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
(45.0)^2&=0+2(4.50)\Delta x\\
9.00\Delta x&=2030\\
\Delta x&={2030\over 9.00}\\
\Delta x&=226\,\mathrm{m}
\end{align*}
