Answer:
[tex]1.0\; \rm m \cdot s^{-1}[/tex].
Explanation:
Consider a time period of duration [tex]\Delta t[/tex]. Let change in the position of an object during that period of time be denoted as [tex]\Delta x[/tex]. The average velocity of that object during that period would be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}[/tex].
For the toy car in this question, the time interval has a duration of [tex]2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s[/tex]. That is: [tex]\Delta t = 1.5\; \rm s[/tex]. (One decimal place, two significant figures.)
On the other hand, what would be the change in the position of this toy car during that [tex]1.5\; \rm s[/tex]?
Note, that from readings on the snapshot in the diagram:
- The position of the toy car was [tex]0.1\; \rm m[/tex] at [tex]t = 0.5\; \rm s[/tex] (the beginning of this [tex]1.5[/tex]-second time period.)
- The position of the toy car was [tex]1.6\; \rm m[/tex] at [tex]t = 2.0\; \rm s[/tex] (the end of this [tex]1.5[/tex]-second time period.)
Therefore, the change to the position of this toy car over that time period would be [tex]\Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m[/tex]. (One decimal place, two significant figures.)
The average velocity of this car over this period of time would thus be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}[/tex]. (Two significant figures.)
