There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:
Consider an athlete running a 40-m dash. The position of the athlete is given by [tex]d(t)=\frac{t^{2}}{6}+4t[/tex] where d is the position in meters and t is the time elapsed, measured in seconds.
Compute the average velocity of the runner over the intervals:
(a) [1.95, 2.05]
(b) [1.995, 2.005]
(c) [1.9995, 2.0005]
(d) [2, 2.00001]
Answer
(a) 6.00041667m/s
(b) 6.00000417 m/s
(c) 6.00000004 m/s
(d) 6.00001 m/s
The instantaneous velocity of the athlete at t=2s is 6m/s
Step by step Explanation:
In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:
[tex]V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}[/tex]
so let's take the first interval:
(a) [1.95, 2.05]
[tex]V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}[/tex]
we get that:
[tex]d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125[/tex]
[tex]d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167[/tex]
so:
[tex]V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s[/tex]
(b) [1.995, 2.005]
[tex]V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}[/tex]
we get that:
[tex]d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831[/tex]
[tex]d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835[/tex]
so:
[tex]V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s[/tex]
(c) [1.9995, 2.0005]
[tex]V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}[/tex]
we get that:
[tex]d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358[/tex]
[tex]d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358[/tex]
so:
[tex]V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s[/tex]
(d) [2, 2.00001]
[tex]V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}[/tex]
we get that:
[tex]d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333[/tex]
[tex]d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333[/tex]
so:
[tex]V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s[/tex]
Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s
