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If a ball is thrown in the air with a velocity $34$ ft/s, its height in feet $t$ seconds later is given by

$$ y = 34 t − 16 t^2 .$$

Find the average velocity for the time period beginning when $t = 2$ and lasting $0.5$ second, $0.1$ second, $0.05$ second, $0.01$ second and estimate the instantaneous velocity when $t = 2$.

I tried to solve by doing the following:

$y=34(2)-16(2)^2$

$y=68-64$

$y=a$

$4/0.5=8$ft/s

but I was told that answer is incorrect.

What did I do wrong?

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  • $\begingroup$ This is a limit approximation. If we say $y=f(x)$, then we can write that the first average is ${f(2.5)-f(2)\over 0.5}={-15-4\over 0.5}=-38$. $\endgroup$ – abiessu Sep 18 '13 at 21:50
  • $\begingroup$ How did you get -15 for f(2.5) and 4 for f(2) ? $\endgroup$ – Grey Sep 18 '13 at 22:22
  • $\begingroup$ I used $f(2.5) = 34(2.5)-16(2.5)^2 = 85-100=-15$ and $f(2)=34(2)-16(2)^2=68-64=4$. I saw the $62$ in the second term of your original post before it was edited and that $16\cdot 4$ is not $62$... $\endgroup$ – abiessu Sep 18 '13 at 22:26
  • $\begingroup$ Thanks. And how to get get the instantaneous velocity when t = 2 ? $\endgroup$ – Grey Sep 18 '13 at 22:40
  • $\begingroup$ That is the point of the exercise, first calculate $f(2.5)-f(2)\over 0.5$, then $f(2.1)-f(2)\over .1$, and so on, and use this series of values to estimate the instantaneous velocity at $t=2$. $\endgroup$ – abiessu Sep 18 '13 at 22:42
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You are being asked for the average velocity over the time span $2$ to $2.5$ seconds, among others. To do that one, you need $y(2.5)$ and $y(2)$ Then the average velocity in that span is $\frac {y(2.5)-y(2)}{2.5-2}$

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There is enough information here to complete the following process:

Given $y(t)=34t-16t^2$, we have:

$$y(2)=34\cdot 2-16\cdot 4=68-64=4$$

$$y(2.5)=34\cdot 2.5-16\cdot 2.5^2=85-100=-15$$

$$y(2.1)=34\cdot 2.1-16\cdot 2.1^2=71.4-70.56=0.84$$

$$y(2.05)=2.46$$ (using calculator)

$$y(2.01)=3.6984$$ (using calculator)

Now, the remaining task is to calculate $-19\over 0.5$, $-3.16\over 0.1$, $-1.54\over 0.05$, and $-0.3016\over 0.01$ and see if these numbers are approaching a particular number. I got each of these fractions by writing $y(2+t)-y(2)\over t$ for each of the values $t\in \{0.5,0.1,0.05,0.01\}$.

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