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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height in feet after t seconds is given by y=30t-10t^2. a) Find the average velocity over the time interval [2,2+ h] where h is given as follows: h= 0.1 ______ ft/s h= 0.01 _____ ft/s h= 0.001 _____ ft/s Estimate the instantaneous velocity when t= 2 _____ ft/s

User Dattatray
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Answer:

a) h = 0.1:
\bar v = -11\,(ft)/(s), h = 0.01:
\bar v = -10.1\,(ft)/(s), h = 0.001:
\bar v = -10\,(ft)/(s), b) The instantaneous velocity of the ball when
t = 2\,s is
-10 feet per second.

Explanation:

a) We know that
y = 30\cdot t -10\cdot t^(2) describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (
\bar v), measured in feet per second, can be done by means of the following definition:


\bar v = (y(2+h)-y(2))/(h)

Where:


y(2) - Position of the ball evaluated at
t = 2\,s, measured in feet.


y(2+h) - Position of the ball evaluated at
t =(2+h)\,s, measured in feet.


h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:


h = 0.1\,s


y(2) = 30\cdot (2) - 10\cdot (2)^(2)


y (2) = 20\,ft


y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^(2)


y(2.1) = 18.9\,ft


\bar v = (18.9\,ft-20\,ft)/(0.1\,s)


\bar v = -11\,(ft)/(s)


h = 0.01\,s


y(2) = 30\cdot (2) - 10\cdot (2)^(2)


y (2) = 20\,ft


y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^(2)


y(2.01) = 19.899\,ft


\bar v = (19.899\,ft-20\,ft)/(0.01\,s)


\bar v = -10.1\,(ft)/(s)


h = 0.001\,s


y(2) = 30\cdot (2) - 10\cdot (2)^(2)


y (2) = 20\,ft


y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^(2)


y(2.001) = 19.99\,ft


\bar v = (19.99\,ft-20\,ft)/(0.001\,s)


\bar v = -10\,(ft)/(s)

b) The instantaneous velocity when
t = 2\,s can be obtained by using the following limit:


v(t) = \lim_(h \to 0) (x(t+h)-x(t))/(h)


v(t) = \lim_(h \to 0) (30\cdot (t+h)-10\cdot (t+h)^(2)-30\cdot t +10\cdot t^(2))/(h)


v(t) = \lim_(h \to 0) (30\cdot t +30\cdot h -10\cdot (t^(2)+2\cdot t\cdot h +h^(2))-30\cdot t +10\cdot t^(2))/(h)


v(t) = \lim_(h \to 0) (30\cdot t +30\cdot h-10\cdot t^(2)-20\cdot t \cdot h-10\cdot h^(2)-30\cdot t +10\cdot t^(2))/(h)


v(t) = \lim_(h \to 0) (30\cdot h-20\cdot t\cdot h-10\cdot h^(2))/(h)


v(t) = \lim_(h \to 0) 30-20\cdot t-10\cdot h


v(t) = 30\cdot \lim_(h \to 0) 1 - 20\cdot t \cdot \lim_(h \to 0) 1 - 10\cdot \lim_(h \to 0) h


v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at
t = 2\,s:


v(2) = 30-20\cdot (2)


v(2) = -10\,(ft)/(s)

The instantaneous velocity of the ball when
t = 2\,s is
-10 feet per second.

User Eden Moshe
by
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