be 1[tex]\bold{_{a}}[/tex]. Distance
= 28 km
Solution:
Given the simple problem, we don't need to use the complex distance formula therefore it can be done as,
[tex]\bold{d = 10km + 8km + 10km}\\\\d = [(10 + 8) + 10]\\\\d = 18 + 10\\\\\boxed{d = 28}[/tex]
thus, Jean covered a distance of 28km
1[tex]\bold{_{b}}[/tex]. Displacement
= 12 km
Solution:
Since the problem illustrates a 1-dimensional motion, therefore, Jean's motion can be illustrated as;
---north 10km------->
<--south 8km----
---north 10km------->
thus,
[tex]\begin{document} \begin{tabular}{lS} & 10km\\ &-8km\\ $+${data-answer}amp; 10km\\\rule{16pt}{1pt} \\ & 12km\\ \end{tabular}[/tex]
(I am unable to place the bottom line due to some perplexing circumstances, hopefully, is lucid)
2[tex]\bold{_{a}}[/tex]. Distance
= 7 km
Solution:
Since distance (d) is the overall 1-dimensional covered distance walked by John, therefore adding is applicable
[tex]\bold{d = 3 km + 4km}\\\\d = 3 km + 4km\\\\\boxed{d = 7km}[/tex]
thus, John covered a distance of 7km
2[tex]\bold{_{b}}[/tex]. Displacement
= 5 km
Formula:
Derived from the Pythagorean theorem, [tex]a^{2} + b^{2} = c^{2}[/tex]
Where,
a - Side of the right triangle
b - Side of the right triangle
c - Hypotenuse
therefore, the magnitude of the displacement vector can be written as;
[tex]d = [(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]^{\frac{1}{2}}[/tex]
Solution:
[tex]\bold{d = [(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]^{\frac{1}{2}}}\\\\d = [(0 - 4)^{2} + (3 - 0)^{2}]^{\frac{1}{2}}\\\\d = [(-4)^{2} + (3)^{2}]^{\frac{1}{2}}\\\\d = (16 + 9)^{\frac{1}{2}}\\\\d = (25)^{\frac{1}{2}}\\\\\boxed{d = 5}[/tex]
Definition (Distance):
- Refers to the length between two points.
- Refers to the 1-dimensional coverage of the subject during its motion.
Definition (Displacements):
- Refers to the subject's shortest distance from its starting point to its final position which has undergone motion.
An image is shown to give a better understanding of what Distance and Displacement is
