All Solutions
Page 42: Section Review
textbf{underline{textit{Solution}}}
$$
And checking the given diagram carefully we notice that the points are equidistant which means that the baby is moving with a constant pace having a uniform speed.\
And the distance between 2 consecutive dots is 20 cm, thus the baby crawls a distance of 20 cm in 1 second, therefor the following textbf{table} describes the particle diagram
begin{center}
begin{tabular}{||c||c| c | c| c| c| c| c| c| c||}
hline hline
Position $left(rm{cm} right)$&0 & 20 & 40 & 60 & 80 & 100 & 120 &140 & 160\
hline
Time (s)& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\
hline hline
end{tabular}
end{center}
textbf{underline{textit{Solution}}}
$$
$$
m= dfrac{y_2-y_1}{x_2-x_1}tag{1}
$$
Where using any 2 points on the line and equation (1) we find that the slope of the straight line is 20 m/s, hence the hockey puck glides with a constant speed of 20 m/s.
Thus we can model such motion as equidistant dots “as the hockey pucks glides with a constant speed” and the distance between two successive points is 20 m such that the time interval between two consecutive dots is 1 seconds.
Therefor particle model $textbf{diagram}$ is as follows
textbf{underline{textit{Solution}}}
$$
Where we find that the $x$-coordinate of the point of intersection line is 0.5 s, thus the hockey puck was 10 m beyond the origin exactly at 0.5 sec.
$$
m=dfrac{y_2-y_1}{x_2-x_1}
$$
Is 20 m/s, and since the straight line is passing through the origin therefor the equation of this straight line is
$$
y=20 x tag{1}
$$
And since we want to find the exact moment at which the hockey puck is at a position 10 m beyond the origin, thus we substitute in the equation (1) by $y=10$ and calculate $x$, we get
$$
begin{align*}
10 &= 20 x\
x&= dfrac{10}{20} \
&= fbox{$0.5~ rm{s}$}
end{align*}
$$
Thus, the hockey goes 10 m beyond the origin at 0.5 seconds.
textbf{underline{textit{Solution}}}
$$
Where we find that the $y$-coordinate of the point of intersection line is at a position 100 m beyond the origin, while at time 0 seconds the hockey puck was at the origin, thus in a time interval of 5 seconds the hockey puck cuts a distance of 100 m.
[ m=dfrac{y_2-y_1}{x_2-x_1}]
Is 20 m/s, and since the straight line is passing through the origin therefor the equation of this straight line is
begin{align*}
y&=20 x tag{1}
end{align*}
And since we want to find the exact moment at which the hockey puck goes beyond 10 m, thus we substitute in the equation (1) by $x=0$ and $x=5$ and calculate $y$ at each time, we get
begin{align*}
intertext{At time $x=0$}
y_0 &= 20 x\
y_0&= 0 \
intertext{At time $x=5$}
y_5 &= 20 times 5\
&= 100 ~ rm{m}
intertext{Thus, the hockey in a time interval between 0 and 5 seconds cuts a distance }
d&=y_5 – y_0\
&= 100-0\
&= 100 ~ rm{m}
end{align*}
Thus, the hockey puck cuts a distance of 100 m in a time interval of 5 seconds.
The puck was at position $d_f=80m$ at time $t_f=4.0s$
The time it took to change position from $d_i$ to $d_f$ is
$$
Delta t=t_f-t_i=2.0s
$$
We need to state whether they describe the same movement.
On the particle model, we see $8$ dots in total, so we have $7$ time intervals. We are told that each interval corresponds to $2text{ s}$.
We will look at the first $5$ time intervals(the first $6$ dots) as we can see that the total distance traveled during these $5$ intervals is $10text{ m}$.
Since we know that $10text{ m}$ is traveled during $5 cdot 2text{ s}=10text{ s}$, we conclude that:
$$
v_1 = frac{10text{ m}}{10text{ s}} = 1 , frac{text{m}}{text{s}}
$$
On the graph, we see that a distance of $12text{ m}$ is traversed during a time interval of $3text{ s}$.
Using this we can find the velocity to equal:
$$
v_2 = frac{12text{ m}}{3text{ s}} = 4 , frac{text{m}}{text{s}}
$$
In the previous two parts, we have found that even though both the particle model and the graph describe bodies moving uniformly, the velocity of the body on the particle model, $v_1$, is much smaller than the velocity corresponding to the body moving on the graph, $v_2$.
