This question is incomplete, the complete question is;
A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.
(a) Write a formula for the average velocity of the ball for t near 4.
(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds
(c) What is your estimate for the instantaneous velocity of the ball at t = 4
Answer:
a)
Average velocity, (Vavg) of the ball for t near 4.
Vavg = [s(4) - s(0)] / (4 - 0)
Where s(4) = -16 × 4² + 8 × √4= - 240 m
s(0) = -16 × 0 + 8 * 0 = 0
b)
duration = 1 sec
Vavg = [s(5) - s(4)] / (5 - 4)
s(5) = -16 × 52 + 8 × √5 = - 382 m
s(4) = -16 × 42 + 8 √4 = - 240 m
Vavg = (-382 - (-240)) / (5 - 4)
Vavg = - 142.1 m/s
duration = 0.5 sec
Vavg = [s(4.5) - s(4)] / (4.5 - 4)
s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m
s(4) = -16 × 42 + 8 × √4 = - 240 m
Vavg = (-307 - (-240)) / (4.5 - 4)
Vavg= - 134.1 m/s
duration = 0.05 sec
Vavg = [s(4.05) - s(4)] / (4.05 - 4)
s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m
s(4) = -16 × 42 + 8 × √4 = - 240 m
Vavg = (-246 - (-240)) / (4.05 - 4)
Vavg= - 126.8 m/s
c)
Instantaneous velocity, v = ds/dt
= - 16 × 2 × t + 8 ×× (0.5 / √t )
= - 32 × t + 4/√t
ds/dt at t = 4 is,
v = - 32 × 4 + 4 / √4
= - 126 m/s