The boy decided to ride the terror tower. at its maximum height, the ride reaches 65m above 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 model: h=65-4.9t^2 where t is in seconds and h in metres.
a)find the average velocity of the gum on the intervals 2<t<3 and 2<t<2.1
b)find the approximate instantaneous velocity when t=2
*solve using math, not physics.
(a) (h(3)-h(2))/(3-2) and (h(2.1)-h(2))/(2.1-2)
(b) dh/dt = -9.8t so plug in t=2
To find the average velocity of the gum on the given intervals, we need to find the displacement (change in height) of the gum and divide it by the time elapsed in each interval.
a) Average Velocity on the interval 2 < t < 3:
To calculate the average velocity on this interval, we need to find the displacement of the gum between t = 2 and t = 3. We can do that by substituting these values into the given model.
At t = 2: h = 65 - 4.9(2^2) = 65 - 4.9(4) = 65 - 19.6 = 45.4 m
At t = 3: h = 65 - 4.9(3^2) = 65 - 4.9(9) = 65 - 44.1 = 20.9 m
The displacement Δh on this interval is the difference between the final and initial heights:
Δh = 20.9 - 45.4 = -24.5 m (negative because the gum is moving downward)
The time elapsed Δt on this interval is 3 - 2 = 1 s
Average velocity = Δh / Δt = -24.5 m / 1 s = -24.5 m/s (negative indicating downward motion)
b) Average Velocity on the interval 2 < t < 2.1:
Following the same procedure as above:
At t = 2: h = 65 - 4.9(2^2) = 45.4 m
At t = 2.1: h = 65 - 4.9(2.1^2) = 65 - 4.9(4.41) = 65 - 21.6099 = 43.3901 m
The displacement Δh on this interval is the difference between the final and initial heights:
Δh = 43.3901 - 45.4 = -2.0099 m (negative indicating downward motion)
The time elapsed Δt on this interval is 2.1 - 2 = 0.1 s
Average velocity = Δh / Δt = -2.0099 m / 0.1 s = -20.099 m/s
b) Instantaneous Velocity at t = 2:
To find an approximate instantaneous velocity at t = 2, we can calculate the average velocity on a very short interval centered around t = 2. Let's say Δt is very close to zero and take two points, t = 2 - Δt and t = 2 + Δt, as our interval.
Using the same model, we can find two heights:
At t = 2 - Δt: h = 65 - 4.9((2 - Δt)^2)
At t = 2 + Δt: h = 65 - 4.9((2 + Δt)^2)
Next, calculate the average velocity on this interval:
Average velocity = (h(2 + Δt) - h(2 - Δt)) / (2Δt)
Finally, take the limit as Δt approaches zero to get the approximate instantaneous velocity.
Since this involves calculus, the exact solution requires math and physics knowledge. If you're interested in a purely mathematical approach, you may consider using numerical methods like finite differences to approximate the instantaneous velocity at t = 2.