Answer:
a) [tex]v(2.5\,s) = 40\,\frac{m}{s}[/tex], b) [tex]v(3.5\,s) = -60\,\frac{m}{s}[/tex], c) [tex]v(4.5\,m) = 0\,\frac{m}{s}[/tex], d) [tex]v(7.5\,s) = 60\,\frac{m}{s}[/tex]
Explanation:
Mathematically speaking, the instantaneous velocity is the slope of the curve, which can be estimated by means of the definition of secant line, equivalent to the definition of average velocity:
[tex]v(t) = \frac{x(t+0.5\,s)-x(x-0.5\,s)}{(t+0.5\,s)-(t-0.5\,s)}[/tex] (1)
Where:
[tex]x(t+0.5\,s)[/tex], [tex]x(t-0.5\,s)[/tex] - Position of the particle at [tex]t +0.5\,s[/tex] and [tex]t-0.5\,s[/tex], measured in meters.
[tex]t[/tex] - Time, measured in seconds.
Now we proceed to calculate each instantaneous velocity:
a) [tex]t = 2.5\,s[/tex]
[tex]v(2.5\,s) = \frac{x(3\,s)-x(2\,s)}{3\,s-2\,s}[/tex]
[tex]v(2.5\,s) = \frac{120\,m-80\,m}{1\,s}[/tex]
[tex]v(2.5\,s) = 40\,\frac{m}{s}[/tex]
b) [tex]t = 3.5\,s[/tex]
[tex]v(3.5\,s) = \frac{x(4\,s)-x(3\,s)}{4\,s-3\,s}[/tex]
[tex]v(3.5\,s) = \frac{60\,m-120\,m}{1\,s}[/tex]
[tex]v(3.5\,s) = -60\,\frac{m}{s}[/tex]
c) [tex]t = 4.5\,s[/tex]
[tex]v(4.5\,s) = \frac{x(5\,s)-x(4\,s)}{5\,s-4\,s}[/tex]
[tex]v(4.5\,s) = \frac{60\,m - 60\,m}{1\,s}[/tex]
[tex]v(4.5\,m) = 0\,\frac{m}{s}[/tex]
d) [tex]t = 7.5\,s[/tex]
[tex]v(7.5\,s) = \frac{x(8\,s)-x(7\,s)}{8\,s-7\,s}[/tex]
[tex]v(7.5\,s) = \frac{0\,m - (-60\,m)}{1\,s}[/tex]
[tex]v(7.5\,s) = 60\,\frac{m}{s}[/tex]
