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

Question

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.

Answer 1

The 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

Answer 2

from 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

Answer 3

Elise 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?