Question

A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(t), in feet, at different times, t, in seconds: f(t) = −16t2 + 48t + 160 The average rate of change of f(t) from t = 3 seconds to t = 5 seconds is _____ feet per second.

Answer 1

THE RIGHT ANSWER IS -80 I ONLY GOT 1 QUESTION WRONG AND IT WASN"T THIS ONE SO THE ANSWER IS -80 THE PERSON ABOVE EXPLAINS WHY!

Answer 2

The Average rate of change of f(t) from t = 3 seconds to t = 5 seconds is - $v_{avg} = -80\;ft/s$

We have a equation -

$f(t) = -16t^{2} +48t+160$

that calculates the height of the ball from sea level in feet at different times.

We have to find the average rate of change of f(t) from t = 3 seconds to t = 5 seconds.

What is Average velocity?

The rate of change of displacement per unit time is called average velocity. Mathematically -

$v_{avg} = \frac{d_{2} - d_{1} }{t_{2} - t_{1} }$

We have -                      

$f(t) = -16t^{2} +48t+160$

For time t = 3 seconds, f(t) is -

$f(3) = -16(3)^{2} + 48(3) + 160\\f(3) = -144 +144 +160\\f(3) = 160\; feets$

For time t = 5 seconds, f(t) is -

$f(5) = -16(5)^{2} + 48(5) + 160\\f(5) = -400+240+160\\f(5) = -400 + 400\\f(5) = 0 feet$

The formula for average velocity is -

$v_{avg} = \frac{d_{2} - d_{1} }{t_{2} - t_{1} }$

According to question -

$t_{2} = 5s$    and     $t_{1} = 3s$

$d_{2} = f(5) = 0 feet$     and       $d_{1} = f(3)=160$

Substituting the values in the formula of average velocity, we get -

$v_{avg} = \frac{0 - 160}{5 - 3} \\v_{avg} = \frac{-160}{2\\} \\v_{avg} = -80\;ft/s$

Hence, average rate of change of f(t) from t = 3 seconds to t = 5 seconds is $v_{avg} = -80\;ft/s$.

To solve more questions on calculating the average rate of change of height with time, visit the following link -

https://brainly.com/question/22257105

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