Take the code from the Master..
the interval [a,b]
(((13*b)-(1.86*b^2))-((13*a)-(1.86*a^2)))/b-a
make b smaller and smaller until you reach the correct instantaneous velocity
(a) Find the average velocity over the given time intervals.
the interval [a,b]
(((13*b)-(1.86*b^2))-((13*a)-(1.86*a^2)))/b-a
make b smaller and smaller until you reach the correct instantaneous velocity
Let's consider two time intervals:
Interval 1: From t = 0 to t = 2 seconds
Interval 2: From t = 0 to t = 5 seconds
For interval 1, we need to calculate the change in position (Δy) from t = 0 to t = 2 seconds:
Δy = y(2) - y(0)
= (13(2) - 1.86(2)^2) - (13(0) - 1.86(0)^2)
= 26 - 7.44 - 0
= 18.56 meters
The duration of interval 1 is 2 seconds. Therefore, the average velocity (v1) over this interval is:
v1 = Δy / Δt
= 18.56 / 2
= 9.28 m/s (rounded to two decimal places)
For interval 2, we need to calculate the change in position (Δy) from t = 0 to t = 5 seconds:
Δy = y(5) - y(0)
= (13(5) - 1.86(5)^2) - (13(0) - 1.86(0)^2)
= 65 - 46.5 - 0
= 18.5 meters
The duration of interval 2 is 5 seconds. Therefore, the average velocity (v2) over this interval is:
v2 = Δy / Δt
= 18.5 / 5
= 3.70 m/s (rounded to two decimal places)
So, the average velocity over the given time intervals is 9.28 m/s for the first 2 seconds and 3.70 m/s for the first 5 seconds.
For the given time interval, you need to find the displacement at the beginning and the end of this interval. Then, you can calculate the average velocity.
Let's say the initial time is t1 and the final time is t2. The displacement at time t1 is y1=13t1−1.86t1^2, and the displacement at time t2 is y2=13t2−1.86t2^2.
The average velocity can be calculated using the following formula:
Average Velocity = (Displacement)/(Time Interval)
Time Interval = t2 - t1
Using this information, you can now plug in the values to calculate the average velocity.