$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$

where $x_{i}$ and $x_{f}$ are the initial and final position, respectively. This simple linear equation has more to it than just giving you the final position. In fact, $textbf{provided that the velocity and the initial position of an object are known, the position-time equation gives the object’s position $x$ at any time $t$.}$

$$
y=ax+b
$$

where $a$ is the slope, and $b$ is the $y$ intercept.

Looking at the equation of motion

$$
x_{f}=x_{i}+vt
$$

it’s obvious that it’s actually the same as the one above, but with rearranged terms and different names of variables. The slope of the line $v$ represents velocity, and the $y$ intercept represents the initial position $x_{i}$.

$$
v_{1}=2.5 frac{text{m}}{text{s}}
$$

$$
v_{2}=-3.5 frac{text{m}}{text{s}}
$$

$textbf{Knowing that speed is the magnitude (or absolute value) of velocity, we conclude that the second bike has the greater speed.}$

$$
begin{align*}
x_{1}&=-4 text{m}\
x_{2}&=6.7 text{m}\
end{align*}
$$

As discussed before, distance has no association whatsoever with direction. Therefore, the distance between the bicycles at $t=0$ is

$$
begin{align*}
d&=4 text{m}+6.7 text{m}\
&=quadboxed{10.7 text{m}}\
end{align*}
$$

If this is confusing, think of it this way: the first bicycle is at $x_{1}=-4 text{m}$, and the second is at $x_{2}=6.7 text{m}$. The distance from the first bike to the origin (where $x=0$) is $4 text{m}$, and the distance from the origin to the second bicycle is $6.7 text{m}$. Thus, the distance between the two bikes is those two distances added, which $10.7 text{m}$.

$$
begin{align*}
x_{1}&=-1.5 text{m}\
x_{2}&=3.2 text{m}\
end{align*}
$$

The distance between them decreases because they travel in opposite directions, or, towards one another. In this case, the distance between them is

$$
begin{align*}
d&=1.5 text{m}+3.2 text{m}\
&=quadboxed{4.7 text{m}}\
end{align*}
$$

$circ$ The distance at $t=1 text{s}$: $boxed{4.7 text{m}}$

$$
begin{align*}
x&=3.0 text{m}-5.0 frac{text{m}}{text{s}}times 1.5 text{s}\
&=3.0 text{m}-7.5 text{m}\
&=quadboxed{-4.5 text{m}}\
end{align*}
$$

$$
begin{align*}
x_{f}=x_{i}+vt
end{align*}
$$

we can easily see that the velocity of each bumper car is

$$
begin{align*}
v_{1}&=1.5 dfrac{text{m}}{text{s}}\
v_{2}&=-2.5 dfrac{text{m}}{text{s}}\
end{align*}
$$

Despite having a negative velocity, $textbf{the second car has the greater speed, because speed is the magnitude (or absolute value) of velocity,}$

$$
begin{align*}
boxed{abs{v_{2}}>abs{v_{1}}}
end{align*}
$$

$$
begin{align*}
x_{1}&=x_{2}\
x_{i,1}+v_{1}t&=x_{i,2}+v_{2}t\
end{align*}
$$

Plugging in the values and solving for $t$ gives:

$$
begin{align*}
-4.0 text{m}+left(1.5 dfrac{text{m}}{text{s}}right)t&=8.8 text{m}-left(2.5 dfrac{text{m}}{text{s}}right)t\
left(1.5 dfrac{text{m}}{text{s}}+2.5 dfrac{text{m}}{text{s}}right)t&=8.8 text{m}+4.0 text{m}\
rightarrowquad t&=dfrac{12.8 text{m}}{4.0 frac{text{m}}{text{s}}}\
&=quadboxed{3.2 text{s}}\
end{align*}
$$