math
posted by Anonymous on .
If a rock is thrown upward on the planet Mercury with a velocity of 11 m/s, its height in meters t seconds later is given by y = 11t  1.86t sqaure
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.

The height has been given in terms of t (in seconds) as:
height=y(t)=11t1.86t²
To evaluate the average velocity in a given time interval [t1,t2], you would calculate:
(y(t2)y(t1))/(t2t1)
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))/(20)=8.64 m s^{1} 
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