The cubic curve y = x'3 + ax'2 + bx + c passes through the point (1,3) and has tangent line y = x - 2 at the point (0,-2). What
1 answer:
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Answer:
- -108.26
- -108.13
- -108.052
- -108.026
- -108
Step-by-step explanation:
A graphing calculator or spreadsheet is useful for making the repeated function evaluations required.
The average velocity on the interval [a,b] will be ...
v avg = (y(b) - y(a))/(b-a)
Here, all the intervals start at a=3, so the average velocity for the given values of t will be ...
v avg = (y(3+t) -y(3))/((3+t) -3) = (y(3+t) -y(3))/t
This can be computed for each of the t-values given. The results are shown in the attached table.
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We note that the fractional part of the velocity gets smaller in proportion to t getting smaller. We expect it to go to 0 when t goes to 0.
The estimated instantaneous velocity is -108 ft/s.
_____
We can simplify the average velocity equation to ...
v avg = ((48(3+t) -26(t+3)^2) -(48(t+3) -26(3)^2)) / t
= (48t -26(t^2 +6t))/t
= 48 -26t -156
<em> v avg = -108 -26t</em>
Then the average velocity at t=0 is -108.
