Assuming that by "16t2" you mean [tex] 16t^2 [/tex]:
First of all, let's compute the position at the two required times:
[tex] d(2)=16\cdot 2^2 = 16*4 = 64 [/tex]
[tex] d(5)=16\cdot 5^2 = 16*25 = 400 [/tex]
So, the ball traveled a total distance of [tex] \Delta d = 400-64 = 336 [/tex], during the time [tex] \Delta t = 5-2 = 3 [/tex].
The average rate of change is the ratio between these two:
[tex] \frac{\Delta d}{\Delta t} = \frac{336}{3} = 112 [/tex]
Average rate of change of a function d(t) between t1 and t2 can be calculated by
average rate of change
= (d(t2)-d(t1)) /(t2-t1)
Geometrically, it represents the secant between points (t1, d(t1)), and (t2, d(t2)).
Physically, it represents the total displacement between t1 and t2, divided by the elapsed time, t2-t1.
For the given problem,
t1=2
t2=5
t2-t1=5-2=3 seconds
d(t1)=d(2)=16(2^2) = 64 ft
d(t2)=d(5)=16(5^2) = 400 ft
d(t2)-d(t1) = 400-64 = 336 ft
Average rate of change (of displacement)
= (d(t2)-d(t1)) / (t2-t1)
= (336/3)
= 112 ft / s.
