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 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

Answer 1

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.