Question

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 29 ft/s. Its height in feet after t seconds is given by y = 29 t − 22 t 2 . A. Find the average velocity for the time period beginning when t=1 and lasting .01 s: .005 s: .002 s: .001 s: NOTE: For the above answers, you may have to enter 6 or 7 significant digits if you are using a calculator.

Answer 1

Answer:

$\overline{v}_{@\Delta t=0.01s}=-15.22ft/s, \overline{v}_{@\Delta t=0.005s}=-15.11ft/s, \overline{v}_{@\Delta t=0.002s}=-15.044ft/s, \overline{v}_{@\Delta t=0.001s}=-15.022ft/s$

Step-by-step explanation:

Now, in order to solve this problem, we need to use the average velocity formula:

$\overline{v}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}$

From this point on, you have two possibilities, either you find each individual $y_{f}, y_{0}, t_{f}, t_{0}$ and input them into the formula, or you find a formula you can use to directly input the change of times. I'll take the second approach.

We know that:

$t_{f}-t_{0}=\Delta t$

and we also know that:

$t_{f}=t_{0}+\Delta t$

in order to find the final position, we can substitute this final time into the function, so we get:

$y_{f}=29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}$

so we can rewrite our formula as:

$\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-y_{0}}{\Delta t}$

$y_{0}$ will always be the same, so we can start by calculating that, we take the provided function ans evaluate it for t=1s, so we get:

$y_{0}=29t-22t^{2}$

$y_{0}=29(1)-22(1)^{2}$

$y_{0}=7ft$

we can substitute it into our average velocity equation:

$\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-7}{\Delta t}$

and we also know that the initil time will always be 1, so we can substitute it as well.

$\overline{v}=\frac{29(1+\Delta t)-22(1+\Delta t)^{2}-7}{\Delta t}$

so we can now simplify our formula by expanding the numerator:

$\overline{v}=\frac{29+29\Delta t-22(1+2\Delta t+\Delta t^{2})-7}{\Delta t}$

$\overline{v}=\frac{29+29\Delta t-22-44\Delta t-22\Delta t^{2}-7}{\Delta t}$

we can now simplify this to:

$\overline{v}=\frac{-15\Delta t-22\Delta t^{2}}{\Delta t}$

Now we can factor Δt to get:

$\overline{v}=\frac{\Delta t(-15-22\Delta t)}{\Delta t}$

and simplify

$\overline{v}=-15-22\Delta t$

Which is the equation that will represent the average speed of the ball. So now we can substitute each period into our equation so we get:

$\overline{v}_{@\Delta t=0.01s}=-15-22(0.01)=-15.22ft/s$

$\overline{v}_{@\Delta t=0.005s}=-15-22(0.005)=-15.11ft/s$

$\overline{v}_{@\Delta t=0.002s}=-15-22(0.002)=-15.044ft/s$

$\overline{v}_{@\Delta t=0.001s}=-15-22(0.001)=-15.022ft/s$