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2 years ago

What is the average rate of change for h(t) between t=0 and t=2

Mathematics

1 answer:

nordsb [41]2 years ago

6 0

Part a: Option D 18 feet per second

Part b: increasing

Solution:

Height $h(t)=-16 t^{2}+50 t+3$

Part a: To find the average rate of change for h(t) between t = 0 and t = 2.

Substitute t = 0 in h(t).

$h(0)=-16 (0)^{2}+50 (0)+3$

h(0) = 3

Substitute t = 2 in h(t).

$h(2)=-16 (2)^{2}+50 (2)+3$

h(2) = 39

Average rate of change formula:

$$=\frac{h(b)-h(a)}{b-a}$

Here, a = 0 and b = 2.

$$=\frac{h(2)-h(0)}{2-0}$

$$=\frac{39-3}{2}$

$$=\frac{36}{2}$

= 18

Average rate of change = 18 feet per second

Option D is the correct answer.

Part b:

This means height of the ball is increasing for 0 < x < 2.

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