In this problem, we find the time it takes for the vehicle in Guided Example 3.9 to stop. Its initial velocity is $v_text{i} = +11.4~mathrm{m/s}$ and the acceleration is $a = -3.80~mathrm{m/s^{2}}$.

At rest, the final velocity must be $v_text{f} = 0$. We have

$$
begin{align*}
v_text{f} &= v_text{i} + at \
implies t &= frac{v_text{f} – v_text{i}}{a} \
&= frac{0 – 11.4~mathrm{m/s}}{-3.80~mathrm{m/s^{2}}} \
t &= boxed{ 3.00~mathrm{s} }
end{align*}
$$

$$
t = 3.00~mathrm{s}
$$

In this problem, a meteorite of mass $m = 12.25~mathrm{kg}$ strikes a car with velocity $v_text{i} = 130~mathrm{m/s}$. It leaves a dent of $Delta x = 0.22~mathrm{m}$. We find its magnitude of acceleration, assuming that it is constant.

The final velocity must be $v_text{f} = 0$. We have

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2a Delta x \
implies a &= frac{v_text{f}^{2} – v_text{i}^{2}}{2 Delta x} \
&= frac{0 – left( 130~mathrm{m/s} right)^{2}}{2 left( 0.22~mathrm{m} right)} \
&= -38409.09091~mathrm{m/s^{2}} \
a &= -38000~mathrm{m/s^{2}}
end{align*}
$$

The magnitude of this acceleration must be
$$
boxed{ leftvert a rightvert
= 38000~mathrm{m/s^{2}} }
$$

$$
leftvert a rightvert = 38000~mathrm{m/s^{2}}
$$

In this problem, a rocket launches from rest with acceleration $a = 106~mathrm{m/s^{2}}$. We find its speed at height $Delta x = 3.20~mathrm{m}$.

The initial velocity must be $v_text{i} = 0$. We have

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2a Delta x \
v_text{f} &= sqrt{ v_text{i}^{2} + 2a Delta x } \
&= sqrt{0 + 2 left( 106~mathrm{m/s^{2}} right) left( 3.20~mathrm{m} right)} \
&= 26.04611~mathrm{m/s} \
v_text{f} &= boxed{ 26.0~mathrm{m/s} }
end{align*}
$$

$$
v_text{f} = 26.0~mathrm{m/s}
$$

newcommand{tx}[1]{text{#1}}

subsection*{Known}

begin{align}
&v_f^2=v_i^2+2aDelta x, tx{for} a=tx{const}.notag\
&implies a=frac{v_f^2-v_i^2}{2Delta x}
end{align}
subsection*{Calculation}
Givens: $v_i=8.4 frac{tx{m}}{tx{s}}$, $Delta x=7.2 tx{m}$, $v_f=6.4 frac{tx{m}}{tx{s}}$\
From (1) we have:
begin{align*}
a=frac{left(6.4 frac{tx{m}}{tx{s}}right)^2-left(8.4 frac{tx{m}}{tx{s}}right)^2}{2(7.2 tx{m})}=-2.1 frac{tx{m}}{tx{s}^2}
end{align*}

subsection*{Conclusion}

vspace{2pt}
hrule
vspace{1pt}

begin{align*}
boxed{a=-2.1 frac{tx{m}}{tx{s}^2}}
end{align*}
The negative sign indicates that the acceleration is in the opposite direction to the velocity.

$$
begin{align*}
boxed{a=-2.1 frac{text{m}}{text{s}^2}}
end{align*}
$$