All Solutions
Page 47: Section Review
$$
B>D>C>A
$$
We will take into account the negative sign for those objects moving in the negative direction.
$$
begin{align*}
V_{text{A}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{(-3text{ m}) – 3text{ m}}{3text{ s} – 0text{ s}}\
&= – frac{6text{ m}}{3text{ s}}\
&= boxed{- 2 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{align*}
V_{text{B}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{3text{ m} – 0text{ m}}{2text{ s} – 0text{ s}}\
&= frac{3text{ m}}{2text{ s}}\
&= boxed{1.5 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{align*}
V_{text{C}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{(-3text{ m}) – 2text{ m}}{5text{ s} – 0text{ s}}\
&= – frac{5text{ m}}{5text{ s}}\
&= boxed{- 1 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{align*}
V_{text{D}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{4text{ m} – (-1text{ m})}{5text{ s} – 0text{ s}}\
&= frac{5text{ m}}{5text{ s}}\
&= boxed{1 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{gather*}
v_{text{B}} > v_{text{D}} > v_{text{C}} > v_{text{A}}\\
(1.5 > 1 > -1 > -2)
end{gather*}
$$
textbf{underline{textit{Solution}}}
$$
Moreover, we don’t know the scale of the graph is $y$-axis, so instead of stating specifically the initial position of each object numerically we could state the initial position in terms of the graph is unit of scale, i.e. “1 square is a 1 unit of length”.
Also, we will take the upper part of the $y$-axis to be positive, i.e. the value of the position is positive, and the lower part is negative.
Thus, The initial position of each object is as follows:
* A: The initial position of this object is at a +ve 3 units of length “3 squares”.
* B: The initial position of this object is at the origin.
* C: The initial position of this object is at a +ve 2 units of length.
* D: The initial position of this object is a a -ve 1 units of length.
Thus, the arrangement of the initial position of the object from the most positive to the most negative is as follows :
A $>$ C $>$ B $>$ D
As the distance is a measure for how far is a point from the origin, it doesn’t matter whether it is positive or negative, i.e. we could say that the distance in our case is the absolute value of the initial position.
While the position is a location of something relative to the origin, so a 2 equidistant objects and a negative directions “one in the upper part and the other in the lower part of the $y$-axis”would have the same distance but not the same position.
Thus, the initial distance for each object is as follows:
* A: the initial distance from origin is 3 units of length.
* B: The initial distance from origin is zero.
* C: The initial distance from origin is 2 units of length.
* D: The initial distance from origin is 1 units of length.
A $>$ C $>$ D $>$ B
* Arranging by initial position: A $>$ C $>$ B $>$ D
* Arranging by initial distance from origin: A $>$ C $>$ D $>$ B
It describes both the direction of the movement and the rate of displacement through time.
It is a **vector** quantity.
It is a **scalar** quantity.
As we have explained, the average speed is the intensity of the average velocity vector.
But some of those equations can be very abstract and mathematically cumbersome.
Once the essence of the phenomenon is understood, the heavy mathematical artillery can be brought in for a detailed analysis.
