In this problem, we find the time it takes for an airplane starting at rest to move with speed $v_text{f} = 77~mathrm{m/s}$ if it has an average acceleration of $a_text{av} = 8.2~mathrm{m/s^{2}}$.

The equation we use must be

$$
begin{align*}
v_text{f} &= v_text{i} = a_text{av}t \
implies t &= frac{v_text{f} – v_text{i}}{a_text{av}} \
&= frac{77~mathrm{m/s} – 0}{8.2~mathrm{m/s^{2}}} \
&= 9.39024~mathrm{s} \
t &= boxed{ 9.4~mathrm{s} }
end{align*}
$$

$$
t = 9.4~mathrm{s}
$$

In this problem, a horse slows from speed $v_text{i} = 11~mathrm{m/s}$ to $v_text{f} = 5.2~mathrm{m/s}$ for a time interval $Delta t = 3.1~mathrm{s}$. We find the average acceleration.

The average acceleration is the change in velocity divided by the time it accelerates. We have

$$
begin{align*}
a_text{av} &= frac{Delta v}{Delta t} \
&= frac{5.2~mathrm{m/s} – 11~mathrm{m/s}}{3.1~mathrm{s}} \
&= -1.87097~mathrm{m/s^{2}} \
a_text{av} &= boxed{ -1.9~mathrm{m/s^2} }
end{align*}
$$

$$
-1.9~mathrm{m/s^2}
$$

In this problem, a train with constant acceleration reaches the speed of $v_{1} = 4.7~mathrm{m/s}$ in $t_{1} = 5.0~mathrm{s}$. We find the speed after another $6.0~mathrm{s}$, which is at time $t_{2} = 11~mathrm{s}$.

The acceleration is constant, and we can get the proportion for the speed at different times.

$$
begin{align*}
v &= a_text{av}t \
v &propto t \
frac{v_{1}}{v_{2}} &= frac{t_{1}}{t_{2}} \
implies v_{2} &= v_{1} frac{t_{2}}{t_{1}} \
&= left( 4.7~mathrm{m/s} right) left( frac{11~mathrm{s}}{5.0~mathrm{s}} right) \
&= 10.34~mathrm{m/s} \
v_{2} &= boxed{ 10~mathrm{m/s} }
end{align*}
$$

$$
v_{2} = 10~mathrm{m/s}
$$