If a ball is thrown into the air with a velocity of 44 ft/s, its height in feet t seconds later is given by y = 44t − 16t2.
(a) find the average velocity for the time period beginning when t = 2 and lasting for each of the following.
(i) 0.5 seconds
(ii) 0.1 seconds
(iii) 0.05 seconds
(iv) 0.01 seconds

Given:
u = 44 ft/s, the initial upward velocity of the ball

The height in feet after t seconds is
y = 44t - 16t²

Initial time is t₁ = 2 s.
The height at 2 s is
h₁ = 44*2 - 16*(2²) = 24 ft

(i) When the duration is 0.5 s, the final time is
    t₂ = 2 +0.5 = 2.5
    The height at 2.5 s is
     h₂ = 44*2.5 - 16(2.5²) = 10 ft
    The average velocity is
    v = (10 - 24 ft)/(0.5 s) = -28 ft/s
    - 28 ft/s

(ii) When the duration is 001s, then t₂ = 2.1 s
    h₂ = 44*2.1 - 16(2.1²) = 21.84 ft
    The average velocity is
    v = (21.84 - 24)/0.1 = -21.6 ft/s
    - 21.6 ft/s

(iii) When the duration is 0.05s, t₂ = 2.05 s
    h₂ = 44*2.05 - 16(2.05²) = 22.96
   Average velocity is
   v = (22.96 - 24)/.05 = - 20.8 ft/s
     - 20.8 ft/s

(iv) When the duration is 0.01 s, t₂ = 2.01 s
  h₂ = 44*201 - 16(2.01²) = 23.7984 ft
   v = (23.7984 - 24)/0.01 = - 20.16 ft/s
  - 20.16 ft/s

As the duration gets smaller, the computed average velocity approaches the true value at 2 s, because it is the derivative of y with respect to t.

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.