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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?

Sagot :

MrRoyal

Answer:

The rate of change is 12ft/s

Step-by-step explanation:

Given

[tex]h(t) = -16t^2 + 20t[/tex]

Required

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

The time she jumps is represented as: t = 0

So, calculate h(0)

[tex]h(t) = -16t^2 + 20t[/tex]

[tex]h(0) = -16 * 0^2 + 20 * 0 = 0[/tex]

At t = 1/2

[tex]h(t) = -16t^2 + 20t[/tex]

[tex]h(1/2) = -16 * 1/2^2 + 20 * 1/2 = 6[/tex]

Rate of change is calculated as:

[tex]Rate = \frac{f(b) - f(a)}{b - a}[/tex]

In this case:

[tex]Rate = \frac{h(b) - h(a)}{b - a}[/tex]

Where

[tex](a,b) = (0,1/2)[/tex]

So, we have:

[tex]Rate = \frac{h(b) - h(a)}{b - a}[/tex]

[tex]Rate = \frac{h(1/2) - h(0)}{1/2 - 0}[/tex]

[tex]Rate = \frac{6 - 0}{1/2 - 0}[/tex]

[tex]Rate = \frac{6}{1/2}[/tex]

[tex]Rate =12[/tex]

The rate of change is 12ft/s

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