Question

a ball is shot straight upward. with it's height, in feet, after t seconds given by the function f(t)=-16t^2+192t. Find the average velocity of the ball from t=1 to t=6

Answer 1

ANSWER

$80 {ms}^{ - 1}$

EXPLANATION

The average velocity of the ball is the rateof displacement over the total time.

The height of the ball, in feet, after t seconds is given by the function:

$f(t)=-16t^2+192t$

At time t=1, the height of the ball is

$f(1)=-16(1)^2+192(1)$

$f(1)=-16+192 = 176ft$

At time t=6, the height of the ball is

$f(6)=-16(6)^2+192(6)$

$f(6)=-16(36)+192(6)$

$f(6)=-576+1152 = 576$

The average velocity

$= \frac{f(6) - f(1)}{6 - 1}$

$= \frac{576- 176}{6 - 1}$

$= \frac{400}{5}$

$= 80 {ms}^{ - 1}$