If a rock is thrown upward on the planet Mercury with a velocity of 11 m/s, its height in meters t seconds later is given by y = 11t - 1.86t sqaure
a : Find the average velocity over the given time intervals: [1, 2] , [1, 1.5] , [1, 1.1] , [1, 1.01] [1, 1.001]
b: Estimate the instantaneous velocity when t = 1.
The height has been given in terms of t (in seconds) as:
height=y(t)=11t-1.86t²
To evaluate the average velocity in a given time interval [t1,t2], you would calculate:
(y(t2)-y(t1))/(t2-t1)
where y(t2)=height at t=t2, and
y(t1)=height at t=t1.
For example, for the interval
[0,2],
height at t=0: y(0)=11(0)-1.18(0)²=0
height at t=2: y(2)=11(2)-1.18(2)²=17.28
Average velocity = (y(2)-y(0))/(2-0)=8.64 m s-1
from 8 feet above a swimming pool susan throws a ball upward with a velocity of 54 feet per second find the maximum height reached by the ball and the time that this height is reached
Elise used 2/9 of piece of string to tie a parcel and another 1/4 of the string to tie a box. What fraction of the string was left?
a) To find the average velocity over a given time interval, we need to calculate the change in height divided by the change in time.
For the interval [1, 2]:
Change in time, Δt = 2 - 1 = 1 second
Change in height, Δy = y(2) - y(1) = (11(2) - 1.86(2)^2) - (11(1) - 1.86(1)^2)
= (22 - 7.44) - (11 - 1.86)
= 14.56 - 9.14
= 5.42 meters
Average velocity, v_avg = Δy / Δt
= 5.42 / 1
= 5.42 m/s
For the other intervals, follow the same steps to find the average velocity:
- For [1, 1.5], Δt = 1.5 - 1 = 0.5 seconds
Δy = y(1.5) - y(1)
- For [1, 1.1], Δt = 1.1 - 1 = 0.1 seconds
Δy = y(1.1) - y(1)
- For [1, 1.01], Δt = 1.01 - 1 = 0.01 seconds
Δy = y(1.01) - y(1)
- For [1, 1.001], Δt = 1.001 - 1 = 0.001 seconds
Δy = y(1.001) - y(1)
Repeat the steps to calculate Δy and then find the average velocity for each interval.
b) To estimate the instantaneous velocity at t = 1, we can calculate the average velocity over a very small time interval centered around t = 1. As the time interval becomes smaller, the average velocity approaches the instantaneous velocity at t = 1.
Let's estimate the instantaneous velocity for a small time interval, say, Δt = 0.001.
At t = 1, y = 11(1) - 1.86(1)^2 = 11 - 1.86 = 9.14 meters
Now, find the height at t = 1 + Δt = 1.001 seconds:
y = 11(1.001) - 1.86(1.001)^2
Calculate the change in height, Δy = y(1.001) - y(1).
Finally, calculate the average velocity over the small time interval: Δy / Δt.
By repeating these calculations for smaller values of Δt, we can have a better estimate of the instantaneous velocity at t = 1.