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 43 ft/s. Its height in feet after t seconds is given byy=43t-23t^{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.
B. Estimate the instanteneous velocity when t=1.

Answer 1

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

Answer 2

The average velocity of ball that is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. at different time intervals are given below

(A) The Average velocity at t =0.01 s is -3.230000 feet/s

    The Average velocity at t =0.005 s is -3.114000 feet/s

    The Average velocity at t =0.002 s is -3.045000 feet/s

    The Average velocity at t =0.001 s is -3.020000 feet/s

(B)  The instantaneous velocity of the ball at t=1 s  is -3.000000 feet /s

Given

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s.

We have to find out  the average velocity for the time period beginning when t=1 and lasting

.01 s:

.005 s:

.002 s:

.001 s:

$\rm let\; h \; be \; the\; height\; of \; the\; ball\; at \; any \; time\; t$

$\rm h(t) = 43t - 23t^2 ..............(1)$

Its height in feet after t seconds is given by equation (1)

$\rm h(1) = 43(1) -23\times1^2 = 43-23 =20 \\Similarly\; height\; at\; different\; times \; given\; can \; be \; calculated\; by \; equation\; (1)\\h(1.01) = 43(1.01)-23(1.01)^2 = 19.9677\\h(1.005)= 43(1.005)-23(1.005)^2= 19.98443\\h(1.002)= 43(1.002)-23(1.002)^2 = 19.99391\\h(1.001)= 43(1.001)-23(1.001)^2 = 19.99698$

By the definition of average velocity we can write

$\rm Averge \; velocity = \dfrac{Total\; Displacement }{Total\; time\; taken }$

Case 1

t = 0.01 s

$\rm Average \; velocity = \dfrac{h(1.01)-h(1)}{1.01-1} = (19.9677-20)/(0.01)=- 3.23 \; feet/s$

Case 2

t = 0.005 s

$\rm Average \; velocity = \dfrac{h(1.005)-h(1)}{1.005-1} = (19.98443-20)/(0.005)= -3.114 \; feet/s$

Case 3

t = 0.002 s

$\rm Average \; velocity = \dfrac{h(1.002)-h(1)}{1.002-1} = (19.99391-20)/(0.002)= -3.045 \; feet/s$

Case 4

t = 0.001 s

$\rm Average \; velocity = \dfrac{h(1.001)-h(1)}{1.001-1} = (19.99698-20)/(0.001)= -3.02 \; feet/s$

(B) The instantaneous velocity is given by equation (2)

$\rm \dfrac{dh}{dt}= 43-46t...........(2)\\So\; instantaneous\; velocity\; at\; t=1 \; s\\h'(1) = 43-46(1) = -3 \; feet /s$

The instantaneous velocity of the ball at t=1 s  is -3.000000 feet /s

For more information please refer to the link given below

https://brainly.com/question/862972