a) plug the values for the ends of the intervals into the height equation
... the change in height over the time interval is the average velocity
b) is calculus allowed?
... dh/dt = -9.8 t
-9.8 m/s^2 is gravitational acceleration
... 9.8 m/s downward for each second of freefall
... after 2 sec, v = 2 * -9.8 m/s
The boy decides to ride the Terror Tower. At its maximum height, the ride reaches 65 m above the ground. When the boy reaches the top of the tower, his gum falls out of his mouth. The height of the gum can be given by the mathematical model h(t) = 65 - 4.9t^2, where t is the time in seconds and h is the height measured in meters.
a) Find the average velocity of the gum on the intervals 2≤t≤3 and 2≤t≤2.1
b) Find the instantaneous velocity when t = 2.
3 answers
So would the answer for part B, be +19.6 or -19.6?
PS thank you for the reply
PS thank you for the reply
Im not clear about the part where you said "the change in height over the time interval is the average velocity"
1 answer