How complex fractions can be used with ratios
1 answer:
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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 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:
so let's take the first interval:
(a) [1.95, 2.05]
we get that:
so:
(b) [1.995, 2.005]
we get that:
so:
(c) [1.9995, 2.0005]
we get that:
so:
(d) [2, 2.00001]
we get that:
so:
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