A ball is thrown vertically upward from the top of a building. The height (in meters) of the ball after t seconds is given by the function s(t) = -(t-3)^2+ 14. Find the instantaneous velocity of the ball at t= 4 seconds by considering the average velocities over the intervals [3.5, 4],
[3.7.4]. [3.9,4]. [3.99, 4], [4, 4.01], [4, 4.1], [4, 4.3], and [4,4.5).
ОА.
1.50 m/sec.
OB.
2.00 m/sec.
Ос.
-2.00 m/sec.
OD. -1.50 m/sec.

9514 1404 393

  C.  -2.00 m/sec

Step-by-step explanation:

The average velocity on the interval [a, b] is found by ...

  m = (s(b) -s(a))/(b -a)

One end of the interval remains constant here, so we can define 'd' so that the interval is [4, 4+d]. Then the average velocity is ...

  m = (s(4 +d) -s(4))/((4 +d) -4)

  m = (s(4+d) -s(4))/d

The attached table shows the average velocity values on the intervals required by the problem statement. Respectively, they are ...

  -1.5 m/s, -1.7 m/s, -1.9 m/s, -1.99 m/s, 2.01 m/s, 2.1 m/s, 2.3 m/s, 2.5 m/s

We expect the instantaneous velocity at d=0 to be the average of the values at d=-0.01 and d=+0.01. We estimate the instantaneous velocity at t=4 seconds to be -2.00 m/s.

2 x t divided by n

step-by-step explanation:

ian gonna lie that is way to much