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.
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
So would the answer for part B, be +19.6 or -19.6?
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"
To find the average velocity on the given intervals, we need to calculate the change in height and divide it by the change in time.
a) Average velocity on the interval 2≤t≤3:
To find the change in height, we subtract the initial height from the final height: h(3) - h(2). Substituting the values of t into the equation for h(t), we have:
h(3) = 65 - 4.9(3)^2 = 35.05 m
h(2) = 65 - 4.9(2)^2 = 55.2 m
Therefore, the change in height is: Δh = h(3) - h(2) = 35.05 - 55.2 = -20.15 m
The change in time is: Δt = 3 - 2 = 1 s
Average velocity is given by: Vavg = Δh / Δt = -20.15 / 1 = -20.15 m/s
b) Average velocity on the interval 2≤t≤2.1:
Using the same process as above, we find:
h(2.1) = 65 - 4.9(2.1)^2 = 54.759 m
Therefore, the change in height is: Δh = h(2.1) - h(2) = 54.759 - 55.2 = -0.441 m
The change in time is: Δt = 2.1 - 2 = 0.1 s
Average velocity is given by: Vavg = Δh / Δt = -0.441 / 0.1 = -4.41 m/s
Now let's move on to finding the instantaneous velocity when t = 2.
To find the instantaneous velocity, we need to calculate the derivative of the position function h(t) with respect to time t.
Taking the derivative of h(t) = 65 - 4.9t^2, we get: h'(t) = -9.8t
Substituting t = 2 into the derivative, we get:
h'(2) = -9.8(2) = -19.6 m/s
Therefore, the instantaneous velocity when t = 2 is -19.6 m/s.