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December 21, 2014

December 21, 2014

Posted by **Anonymous** on Saturday, August 28, 2010 at 6:33pm.

a : Find the average velocity over the given time intervals: [1, 2] , [1, 1.5] , [1, 1.1] , [1, 1.01] [1, 1.001]

b: Estimate the instantaneous velocity when t = 1.

- math -
**MathMate**, Saturday, August 28, 2010 at 6:40pmThe height has been given in terms of t (in seconds) as:

height=y(t)=11t-1.86t²

To evaluate the average velocity in a given time interval [t1,t2], you would calculate:

(y(t2)-y(t1))/(t2-t1)

where y(t2)=height at t=t2, and

y(t1)=height at t=t1.

For example, for the interval

[0,2],

height at t=0: y(0)=11(0)-1.18(0)²=0

height at t=2: y(2)=11(2)-1.18(2)²=17.28

Average velocity = (y(2)-y(0))/(2-0)=8.64 m s^{-1}

- math -
**Anonymous**, Friday, January 21, 2011 at 11:15pmfrom 8 feet above a swimming pool susan throws a ball upward with a velocity of 54 feet per second find the maximum height reached by the ball and the time that this height is reached

- math -
**ruel**, Sunday, February 13, 2011 at 3:32amElise used 2/9 of piece of string to tie a parcel and another 1/4 of the string to tie a box. What fraction of the string was left?

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