In this problem, we find the time it takes for the drag racer in Guided Example 3.8 to reach $x_text{f} = 5.00~mathrm{m}$. It accelerates from rest with rate $a = 7.40~mathrm{m/s^{2}}$. It starts from rest, so $x_text{i} = 0$ and $v_text{i} = 0$.

The equation of motion is

$$
begin{align*}
x_text{f} &= x_text{i} + v_text{i}t + frac{1}{2}at^{2} \
5.00~mathrm{m} &= 0 + 0 + frac{1}{2} left( 7.40~mathrm{m/s^{2}} right) t^{2} \
implies t^{2} &= frac{2 left( 5.00~mathrm{m} right)}{7.40~mathrm{m/s^{2}}} \
t &= sqrt{ frac{2 left( 5.00~mathrm{m} right)}{7.40~mathrm{m/s^{2}}} } \
&= 1.16248~mathrm{s} \
t &= boxed{ 1.16~mathrm{s} }
end{align*}
$$

$$
t = 1.16~mathrm{s}
$$

$$
textbf{Concept :}
$$

The position-time equation for constant acceleration can be used to calculate the displacement of the sled.

$$
textbf{Solution :}
$$

Substituting values in the equation

$$
x_f=x_i+v_i t+frac{1}{2}at^2=0+frac{1}{2}at^2
$$

$$
=0+(1.2m/s)(4.0s)+frac{1}{2}(1.8m/s^2)(4.0s)^2=color{#4257b2} boxed{bf 19m}
$$

$$
x_f=19m
$$

In this problem, a horse accelerates for time $t = 1~mathrm{s}$ and covers a distance $D$. We see what happens to the distance covered when it accelerates for $t = 2~mathrm{s}$. It accelerates from rest, so $x_text{i} = 0$ and $v_text{i} = 0$.

We find the relationship between the covered distance $x_text{f}$ and $t$

$$
begin{align*}
x_text{f} &= x_text{i} + v_text{i}t + frac{1}{2}at^{2} \
&= 0 + 0 + frac{1}{2}at^{2} \
x_text{f} &= left( frac{1}{2}a right)t^{2} \
implies x_text{f} &propto t^{2}
end{align*}
$$

The time of travel is doubled, so the distance $x_text{f}$ must be scaled by $(2)^{2} = 4$. Hence, the new distance covered is
$$
boxed{4D}
$$

$$
4D
$$

In this problem, we are given that a cheetah can accelerate from rest to $v = 25.0~mathrm{m/s}$ in time $t_{1} = 6.00~mathrm{s}$. Assuming that the acceleration is constant, we find the distance it cover in the first $t = 3.00~mathrm{s}$.

First, we find the acceleration

$$
begin{align*}
v &= 0 + at_{1} \
implies a &= frac{v}{t_{1}} \
&= frac{25.0~mathrm{m/s}}{6.00~mathrm{s}} \
a &= frac{25}{6}~mathrm{m/s^{2}}
end{align*}
$$

Now, to find the distance traveled.

$$
begin{align*}
x_text{f} &= x_text{i} + v_text{i}t + frac{1}{2}at^{2} \
&= 0 + 0 + frac{1}{2}left(frac{25}{6}~mathrm{m/s^{2}} right) left( 3.00~mathrm{s} right)^{2} \
&= 18.75~mathrm{m} \
x_text{f} &= boxed{ 18.8~mathrm{m} }
end{align*}
$$

$$
18.8~mathrm{m}
$$