All Solutions
Page 53: Lesson Check
Let’s assume that your friends start the trip at $x=0$ and drive in the positive direction, traversing $40 text{km}$. They noticed that they missed their desired destination by $4 text{km}$ and went back, driving at $20 frac{text{m}}{text{s}}$ in the negative direction. Finally, they get to their destination after driving for 30 minutes in total, and are $36 text{km}$ from the starting point.
During that trip, they drove at $-20 frac{text{m}}{text{s}}$ for some time. However, their average velocity was
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
&=dfrac{x_{f}-x{i}}{Delta t}\
&=dfrac{36 text{km}}{30 text{min}}\
&=dfrac{36 000 text{m}}{1800 text{s}}\
&=quadboxed{20 frac{text{m}}{text{s}}}\
end{align*}
$$
Therefore, your friends are telling the truth when they say that at some point during the trip their velocity was $-20 frac{text{m}}{text{s}}$, even though their average velocity was $20 frac{text{m}}{text{s}}$.
Click to see an example.
$$
begin{align*}
text{average speed}&=dfrac{100 text{m}}{9.58 text{s}}\
&=quadboxed{10.4 frac{text{m}}{text{s}}}\
end{align*}
$$
and in kilometers per hour
$$
begin{align*}
text{average speed}&=10.4 frac{text{m}}{text{s}}cdotleft(dfrac{3.6 frac{text{km}}{text{h}}}{1 frac{text{m}}{text{s}}}right)\
&=quadboxed{37.4 frac{text{km}}{text{h}}}\
end{align*}
$$
begin{align*}
text{in meters per second:} &quadboxed{text{average speed}=10.4 frac{text{m}}{text{s}}}\
\
\
\
text{in kilometers per hour:} &quadboxed{text{average speed}=37.4 frac{text{km}}{text{h}}}\
end{align*}
$$
textbf{Concept:}
$$
The average speed is the distance divided by elapsed time.
$$
textbf{Solution:}
$$
By dividing the distance with time we get:
$$
s=frac{distance}{time}=frac{100m}{9.58s}=10.44m/s=color{#4257b2} boxed{bf 10.4m/s}
$$
Converting this distance to km/hr we multiply be 3600s and divide by 1000m to get the answer:
$$
10.44frac{cancel{m}}{cancel{s}} frac{1km}{1000cancel{m}}frac{3600cancel{s}}{1hr} = color{#4257b2} boxed{bf 37.6 km/hr}
$$
37.6km/hr
$$
$$
begin{align*}
t&=dfrac{2d}{c}\
&=dfrac{2times 3.84cdot 10^{8} text{m}}{3cdot 10^{8} frac{text{m}}{text{s}}}\
&=quadboxed{2.56 text{s}}\
end{align*}
$$
begin{align*}
boxed{t=2.56 text{s}}\
end{align*}
$$
textbf{Concept:}
$$
The time passed is the distance travelled by the average speed. The distance to the Moon is $3.84 times 10^8 m$. We must multiply the distance by two because the signal travels there and back again.
$$
textbf{Solution:}
$$
Calculate the average speed by dividing the distance with speed we get:
$$
t=frac{2d}{s}=frac{2(3.84times 10^8cancel{m})}{3.00times 10^8cancel{m}/s}=color{#4257b2} boxed{bf 2.56s}
$$
t=2.56s
$$
The average speed will be less than$text{color{#4257b2} boxed{bf 25 m/s} as the train spends a longer time driving at the lower speed. The train covers the second 2 km distance in less time at the higher speed than it spent travelling the first 2 km.}$
Calculate the distance travelled first:
$$
textbf{Solution:}
$$
$$
Delta x = 2km + 2km = 4km times 1000 m/km = 4000 m
$$
Now find the time passed for the 1st segment:
$$
Delta t_1 = frac{Delta x_1}{v_1}=frac{2km}{20 m/s}times frac{1000m}{km}=100s
$$
similarly calculate for the second segment we get…
$$
Delta t_2 = frac{Delta x_2}{v_2}=frac{2km}{30 m/s}times frac{1000m}{km}=66.67s
$$
Then average the speed for two segments
$$
v_{av}=frac{Delta x}{Delta t_1+Delta t_2}=frac{4000m}{166.67s}=color{#4257b2} boxed{bf 24m/s}
$$
v_{av}=24m/s
$$
$$
begin{align*}
t_{1}&=dfrac{d_{1}}{v_{1}}\
&=dfrac{2000 text{m}}{20.0 frac{text{m}}{text{s}}}\
&=quadboxed{100 text{s}}\
end{align*}
$$
and the elapsed time during the second distance travelled is
$$
begin{align*}
t_{2}&=dfrac{d_{2}}{v_{2}}\
&=dfrac{2000 text{m}}{30.0 frac{text{m}}{text{s}}}\
&=quadboxed{66.7 text{s}}
end{align*}
$$
Therefore, the average speed is
$$
begin{align*}
text{average speed}&=dfrac{d_{1}+d{2}}{t_{1}+t_{2}}\
&=dfrac{4000 text{m}}{166.7 text{s}}\
&=quadboxed{24.0 frac{text{m}}{text{s}}}\
end{align*}
$$
which means that we have verified the answer to part $textbf{(a)}$.
$textbf{(b)}$ $boxed{text{average speed}=24.0 frac{text{m}}{text{s}}}$
textbf{Concept:}
$$
The average speed can be calculated by diving the distance travelled by the time passed.
$$
textbf{Solution:}
$$
Calculate the distance travelled first:
$$
Delta x = 1km + 1km = 2km
$$
Now find the time passed for the 1st segment:
$$
Delta t_1 = frac{Delta x_1}{v_1}=frac{1cancel{km}}{10 cancel{km}/hr}=0.100h
$$
Similarly calculate for the second segment we get…
$$
Delta t_2 = frac{Delta x_2}{v_2}=frac{1cancel{km}}{30 cancel{km}/hr}=0.033h
$$
Then average the speed for two segments
$$
v_{av}=frac{Delta x}{Delta t_1+Delta t_2}=frac{2km}{0.133hr}=color{#4257b2} boxed{bf 15km/hr}
$$
v_{av}=15km/hr
$$
$$
begin{align*}
t_{1}&=dfrac{s_{1}}{v_{1}}\
&=dfrac{1 text{km}}{10 frac{text{km}}{text{h}}}\
&=0.1 text{h}\
end{align*}
$$
and the time to cover $1 text{km}$ at $30 frac{text{km}}{text{h}}$ is
$$
begin{align*}
t_{2}&=dfrac{s_{2}}{v_{2}}\
&=dfrac{1 text{km}}{30 frac{text{km}}{text{h}}}\
&=0.03 text{h}\
end{align*}
$$
So, the average speed for the total bike ride is
$$
begin{align*}
text{average speed}&=dfrac{text{distance}}{text{elapsed time}}\
&=dfrac{2 text{km}}{0.13 text{h}}\
&=quadboxed{15 frac{text{km}}{text{h}}}\
end{align*}
$$
begin{align*}
boxed{text{average speed}=15 frac{text{km}}{text{h}}}\
end{align*}
$$
