Reading off the graph, we see that the corresponding positions for $t_{i}=0$ and $t_{f}=3 text{s}$ are, respectively, $x_{i}=0$ and $x_{f}=15 text{m}$. The displacement is

$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=15 text{m}-0\
&=quadboxed{15 text{m}}\
end{align*}
$$

Therefore, the correct answer is $textbf{(C)}$.

$$
textbf{(C)}
$$

The slope-intercept form of the line equation is

$$
begin{align*}
y=ax+b\
end{align*}
$$

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

We can also describe motion with the slope-intercept form of the line equation

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

with $x_{i}$ being the intercept and $v$ the slope.

If the velocity of an object is constant in time, then the slope of the line will be constant as well. Therefore, we can conclude that the correct answer is $textbf{(B)}$.

$$
textbf{(B)}
$$

Reading off the graph, we see that the corresponding positions for $t_{i}=0$ and $t_{f}=10 text{s}$ are, respectively, $x_{i}=0$ and $x_{f}=20 text{m}$. The displacement is

$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=20 text{m}-0\
&=quadboxed{20 text{m}}\
end{align*}
$$

Therefore, the correct answer is $textbf{(D)}$.

$textbf{(D)}$

Reading off the graph, we see that the corresponding positions for $t_{i}=8 text{s}$ and $t_{f}=10 text{s}$ are, respectively, $x_{i}=25 text{m}$ and $x_{f}=20 text{m}$. The average velocity is

$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{20 text{m}-25 text{m}}{10 text{s}-8 text{s}}\
&=dfrac{-5 text{m}}{2 text{s}}\
&=quadboxed{-2.5 frac{text{m}}{text{s}}}\
end{align*}
$$

Therefore, the correct answer is $textbf{(B)}$.

$textbf{(B)}$

Up until $t=6 text{s}$ the slope on the graph is positive, which indicates that the object’s velocity is positive. From that moment on, $textbf{the slope is negative, which, in turn, corresponds to a negative velocity, and the negative sign comes from the change in direction of motion.}$ In addition, the slope is steeper when the object moves with a positive velocity, and we know that $textbf{the steeper the slope, the greater the magnitude of velocity, which is speed.}$ So, at $t=6 text{s}$, the object changes both its direction and speed, making $textbf{(A)}$ the correct answer.

If you are unsure about the correct answer, you can explicitly calculate the average velocity for each part of motion.

For the first part, we have:

$$
begin{align*}
v_{avg,1}&=dfrac{Delta x_{1}}{Delta t_{1}}=dfrac{x_{f,1}-x_{i,1}}{t_{f,1}-t_{i,1}}\
&=dfrac{30 text{m}-0}{6 text{s}-0}\
&=quadboxed{5 frac{text{m}}{text{s}}}\
\
abs{v_{avg,1}}&=abs{5 frac{text{m}}{text{s}}}\
&=quadboxed{5 frac{text{m}}{text{s}}}\
end{align*}
$$

and for the second part

$$
begin{align*}
v_{avg,2}&=dfrac{Delta x_{2}}{Delta t_{2}}=dfrac{x_{f,2}-x_{i,2}}{t_{f,2}-t_{i,2}}\
&=dfrac{20 text{m}-30 text{m}}{10 text{s}-6 text{s}}\
&=dfrac{-10 text{m}}{4 text{s}}\
&=quadboxed{-2.5 frac{text{m}}{text{s}}}\
\
abs{v_{avg,2}}&=abs{-2.5 frac{text{m}}{text{s}}}\
&=quadboxed{2.5 frac{text{m}}{text{s}}}\
end{align*}
$$

$textbf{(A)}$

First, let’s write the equation of motion for each runner.

The equation of motion for runner A is

$$
begin{align*}
x_{A}=left(3.0 frac{text{m}}{text{s}}right)times t\
end{align*}
$$

and the equation of motion for runner B is

$$
begin{align*}
x_{B}=20 text{m}-left(2.0 frac{text{m}}{text{s}}right)times t\
end{align*}
$$

When they meet, their positions are the same, so we are solving the equation $x_{A}=x_{B}$ for $t$:

$$
begin{align*}
left(3.0 frac{text{m}}{text{s}}right)times t&=20 text{m}-left(2.0 frac{text{m}}{text{s}}right)times t\
left(3.0 frac{text{m}}{text{s}}+2.0 frac{text{m}}{text{s}}right)times t&=20 text{m}\
rightarrowquad t&=dfrac{20 text{m}}{5.0 frac{text{m}}{text{s}}}\
&=quadboxed{4.0 text{s}}\
end{align*}
$$

To find out their position, we just have to substitute $t$ with $4.0 text{s}$ in either equation.

$$
begin{align*}
x&=3.0 frac{text{m}}{text{s}}cdot 4.0 text{s}\
&=quadboxed{12 text{m}}\
end{align*}
$$

Therefore, the correct answer is $textbf{(D)}$.

$textbf{(D)}$

Let’s look at each option separately.

Option $textbf{(A)}$ is an example of instantaneous velocity.

Option $textbf{(B)}$ is a bit deceitful; the speed of an object is constant, but when moving in a circle, the direction constantly changes; therefore, this isn’t the right answer either.

More or less the same applies for $textbf{(C)}$, speed could be constant during the entire drive, but the direction changes.

This leaves us with $textbf{(D)}$, the speed is constant and the direction of motion remains unchanged, which means that the velocity is also constant.

$textbf{(D)}$

The given equation of motion is analogous to the equation of a straight line written in the slope-intercept form:

$$
begin{align*}
y=b+ax quadLongrightarrowquad x=12text{m}+left(5 frac{text{m}}{text{s}}right)times t\
end{align*}
$$

We can immediately eliminate $textbf{(A)}$ because this option implies that there was a change in velocity; the slope of a straight line is constant which means that velocity is constant, too.

Option $textbf{(B)}$ is obviously not true, as well as option $textbf{(D)}$; acceleration implies a change in velocity.

This leaves us with $textbf{(C)}$: the object moves with a constant velocity of $5 frac{text{m}}{text{s}}$.

$textbf{(C)}$

$textbf{(a)}$      When walking north, the hiker covers

$$
begin{align*}
v_{1}&=dfrac{d_{1}}{t_{1}}\
rightarrowquad d_{1}&=v_{1}t_{1}\
&=2 frac{text{m}}{text{s}}times 30 text{min}cdotdfrac{60 text{s}}{1 text{min}}\
&= quadboxed{3600 text{m}}
end{align*}
$$

and when walking west

$$
begin{align*}
v_{2}&=dfrac{d_{2}}{t_{2}}\
rightarrowquad d_{2}&=v_{2}t_{2}\
&=1 frac{text{m}}{text{s}}times 1 text{h}cdotdfrac{3600 text{s}}{1 text{h}}\
&= quadboxed{3600 text{m}}
end{align*}
$$

The total distance is the sum of these two distances

$$
begin{align*}
d&=d_{1}+d_{2}\
&=3600 text{m}+3600 text{m}\
&=quadboxed{7200 text{m}}\
end{align*}
$$

$textbf{(b)}$      This is a situation when it’s useful to make a sketch of the problem, which is given below.

We know that displacement is the net change in position; i.e., the difference between final and initial position. In this case, we’ll have to calculate the hiker’s displacement as we would calculate the hypotenuse of a right-angled triangle with equal catheti.

Having all this in mind, the displacement is

$$
begin{align*}
(Delta x)^{2}&=(3600 text{m})^{2}+(3600 text{m})^{2}\
&=25 920 000 text{m}^{2}\
\
rightarrowquadDelta x&=sqrt{25 920 000 text{m}^{2}}\
&=5091.168825 text{m}\
&=quadboxed{5091 text{m}}\
end{align*}
$$

When writing the final result, we used $textbf{the rules for mathematical operations for significant figures.}$

$textbf{(c)}$      Distance and displacement are two very different quantities.

$textbf{Distance is a scalar,}$ which means that it has no direction associated with it, it is just a number representing the total length covered during motion, and it’s defined as always being positive.

$textbf{Displacement, however, is a vector,}$ meaning that it has both magnitude and direction. It’s defined as the net change in position, so, relative to the chosen origin and positive direction , it can be positive, negative, or zero.

It’s also important to know that displacement of an object can never be greater than the distance it covered.

$$
begin{align*}
textbf{(a)} quad &boxed{d=7200 text{m}}\
\
textbf{(b)} quad &boxed{Delta x=5091 text{m}}\
\
textbf{(c)} quad &text{Distance is a scalar, whereas displacement is a vector.}\
end{align*}
$$