All Solutions
Page 57: Lesson Check
Along the y axis, we measure position.
$$
textbf{The steepness of the slope tells us more about the average speed: the steeper the slope, the greater the average speed.}
$$
The steepness of the slope tells us more about the average speed: the steeper the slope, the greater the average speed.
$$
begin{align*}
boxed{text{average speed},_{C}<text{average speed},_{A}<text{average speed},_{B}}\
end{align*}
$$
$$
begin{align*}
v_{avg},_{A}&=dfrac{-2 text{m}}{2 text{s}}=-1 frac{text{m}}{text{s}}\
\
\
\
v_{avg},_{B}&=dfrac{2 text{m}}{1 text{s}}=2 frac{text{m}}{text{s}}\
\
\
\
v_{avg},_{C}&=dfrac{-1 text{m}}{2 text{s}}=-0.5 frac{text{m}}{text{s}}\
end{align*}
$$
Therefore, ranking from most negative to most positive velocity, we have
$$
begin{align*}
boxed{v_{avg},_{A}<v_{avg},_{C}<v_{avg},_{B}}\
end{align*}
$$
This could easily have been done without calculation: the only segment that has a positive slope is B, so it's going to rank last. The only question is, which is more negative, A or C? The answer is, again, very simple. Comparing the slopes, you can see that A has a steeper slope, which would correspond to greater (in this case, more negative) velocity.
begin{align*}
textbf{(a)} quad &boxed{text{average speed},_{C}<text{average speed},_{A}<text{average speed},_{B}}\
\
\
\
textbf{(b)} quad &boxed{v_{avg},_{A}<v_{avg},_{C}<v_{avg},_{B}}\
end{align*}
$$
The speed of the tennis player is equal to the magnitude of the slope of the position-time graph. The player has negative average velocity when the position of the player is decreasing where as when it is increasing the player has a positive average velocity.
$$
textbf{Solution:}
$$
The magnitude of the slope of segment B is the largest of the three, and the slope of segment C is the smallest, so we arrive at the ranking: $color{#4257b2} boxed{bf speed C < speed A < speed B}$.
$speed C < speed A < speed B$.
Find the slopes for respective segments starting from segment A we get
$$
textbf{Solution:}
$$
$$
Slope_A=v_{av,A}=frac{Delta x}{Delta t}=frac{-2m}{2.0s}=1.0m/s
$$
Repeat the process for the remaining segment we get
$$
Slope_B=v_{av,B}=frac{Delta x}{Delta t}=frac{2.0m}{1.0s}=2.0m/s
$$
$$
Slope_C=v_{av,C}=frac{Delta x}{Delta t}=frac{-1.0m}{2.0s}=-0.5m/s
$$
By comparison we conclude at the following ranking $color{#4257b2} boxed{bf Velocity_A<Velocity_C<Velocity_B}$
$$
begin{align*}
boxed{text{speed}_{C}<text{speed}_{B}<text{speed}_{A}<text{speed}_{D}}\
end{align*}
$$
$$
begin{align*}
boxed{v_{D}<v_{C}<v_{B}<v_{A}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{text{speed}_{C}<text{speed}_{B}<text{speed}_{A}<text{speed}_{D}}\
\
\
\
textbf{(b)} quad &boxed{v_{D}<v_{C}<v_{B}<v_{A}}\
end{align*}
$$
The speed of the object is equal to the magnitude of the slope of the position-time graph. The object has negative average velocity when the position of the object is decreasing where as when it is increasing the object has a positive average velocity.
$$
textbf{Solution:}
$$
D has the magnitude of slope that is the largest of the four, and the slope of segment C stands smallest, so we arrive at the ranking: $text{color{#4257b2} boxed{bf speed C < speed B < speed A < speed D}.}$
textbf{Concept:}
$$
The speed of the object is equal to the magnitude of the slope of the position-time graph. The object has negative average velocity when the position of the object is decreasing where as when it is increasing the object has a positive average velocity.
$$
textbf{Solution:}
$$
We found that the slopes D and C are negative and slope D have the largest magnitude. Slope A also has a larger magnitude than B. We arrive at the ranking for velocity, from most negative to most positive: $text{color{#4257b2} boxed{bf velocity D < velocity C < velocity B < velocity A}.}$
$$
begin{align*}
v_{avg}=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
end{align*}
$$
Plugging in the values, we obtain the following result
$$
begin{align*}
v_{avg}&=dfrac{2 text{m}-3 text{m}}{5 text{s}-0 text{s}}\
&=quadboxed{-0.2 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
boxed{v_{avg}=-0.2 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
text{smallest speed}&=dfrac{3 text{m}}{2 text{s}}\
&=quadboxed{1.5 frac{text{m}}{text{s}}}\
end{align*}
$$
$textbf{(b)}$ $boxed{text{smallest speed}=1.5 frac{text{m}}{text{s}}}$
Segment B has the smallest slope of the three, so we found that the train has the smallest speed on $text{color{#4257b2} boxed{bf portion B} of its motion.}$
The slope of the segment B gives the speed of the train:
$$
textbf{Solution:}
$$
$$
Slope_B= frac{Delta x_B}{Delta t_B}=frac{-3.0m}{2s}=color{#4257b2} boxed{bf 1.5m/s}
$$
1.5m/s
$$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{5 text{m}-(-3 text{m})}{7 text{s}-0 text{s}}\
&=dfrac{8 text{m}}{7 text{s}}\
&=1.14 frac{text{m}}{text{s}}\
&=quadboxed{1 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
boxed{v_{avg}=1 frac{text{m}}{text{s}}}\
end{align*}
$$
textbf{Concept:}
$$
The slope of the position-time graph is equal to the speed of the train (displacement divided by avg. time)
$$
textbf{Solution:}
$$
Divide the displacement by the time we get:
$$
v_{av}=frac{Delta x}{Delta t}=frac{5.0m-(-3.0m)}{7.0s}=frac{8m}{7s}=color{#4257b2} boxed{bf 1.14m/s}
$$
1.14m/s
$$
