Question

LAN tosses a bone up In the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be represented by the function h(t) = -16t^2 + 20t. What is Spots average rate of ascent, in feet per second, from the time she jumps into the air to the Time she catches the bone at t = 1/2 second?

Answer 1

Answer:

The rate of change is 12ft/s

Step-by-step explanation:

Given

$h(t) = -16t^2 + 20t$

Required

Rate of change from when she jumps till 1/2s

The time she jumps is represented as: t = 0

So, calculate h(0)

$h(t) = -16t^2 + 20t$

$h(0) = -16 * 0^2 + 20 * 0 = 0$

At t = 1/2

$h(t) = -16t^2 + 20t$

$h(1/2) = -16 * 1/2^2 + 20 * 1/2 = 6$

Rate of change is calculated as:

$Rate = \frac{f(b) - f(a)}{b - a}$

In this case:

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

Where

$(a,b) = (0,1/2)$

So, we have:

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

$Rate = \frac{h(1/2) - h(0)}{1/2 - 0}$

$Rate = \frac{6 - 0}{1/2 - 0}$

$Rate = \frac{6}{1/2}$

$Rate =12$

The rate of change is 12ft/s