If a ball is thrown into the air with a velocity of 30m/s, its height in metres after t seconds is given by y=30t-4.9t^2
Fing the average velocity for the time period beginning when t=2 and lasting 1second.
I did V(1)=lim h->0 [[30(1+h)-4.9(1+h)^2]-[30(1)-4.9(1)^2]]/h
After I expanded everything, I am still unable to cancel the h on the bottom.
Thanks in advance.
Note that it said "average velocity", what you tried to do is the instantaneous velocity at a time of t = 1 sec
(it said, "lasting 1 second", meaning there was a duration of 1 second)
so you want the average velocity between 2 and 3 seconds.
height at 2 sec = 30(2) - 4.9(4) = 40.4
height at 3 sec = 30(3) - 4.9(9) = 45.9
so average velocity = (45.9-40.4)/(3-2) = 5.5 m/s
btw, when you expanded your expression you should have arrived at
lim [30 + 30h - 4.9 - 9.8h - 4.9h^2 - 30 + 4.9]/h
= lim[20.2h - 4.9h^2]/h
= lim [20.2 - 4.9h] as h ---> 0
= 20.2
which would be the instantaneous vel at t=1
Well, it seems like you're having a bit of trouble with your calculations. Don't worry, I'm here to help you... or at least try to with some humor!
Let's take a look at your expression and see how we can simplify it.
First, we expand the terms inside the brackets:
[30(1+h)-4.9(1+h)^2] = 30 + 30h - 4.9 - 9.8h - 4.9h^2.
Now, we subtract the initial expression at t=1:
[30(1+h)-4.9(1+h)^2] - [30(1)-4.9(1)^2] = (30 + 30h - 4.9 - 9.8h - 4.9h^2) - 25.1.
Combining like terms, we get:
30 + 30h - 4.9 - 9.8h - 4.9h^2 - 25.1 = 0.9h - 4.9h^2.
Now, let's simplify by dividing everything by h:
(0.9h - 4.9h^2) / h = 0.9 - 4.9h.
Finally, we take the limit as h approaches 0:
lim h->0 (0.9 - 4.9h) = 0.9 - 4.9 * 0 = 0.9.
So, the average velocity for the time period beginning when t=2 and lasting 1 second is approximately 0.9 m/s.
Remember, math can be tricky sometimes, but with a little bit of humor, we can hopefully make it a bit more enjoyable!
To find the average velocity for the given time period, you need to find the change in height during that time period and divide it by the duration.
The equation that describes the height of the ball after t seconds is y = 30t - 4.9t^2.
To find the change in height during the time period from t = 2 to t = 3 (which lasts 1 second), we need to calculate y(3) - y(2).
y(3) = 30(3) - 4.9(3)^2 = 90 - 44.1 = 45.9
y(2) = 30(2) - 4.9(2)^2 = 60 -19.6 = 40.4
The change in height during this time period is y(3) - y(2) = 45.9 - 40.4 = 5.5 meters.
Now, to find the average velocity, we divide the change in height by the duration, which is 1 second.
Average velocity = (change in height) / (duration) = 5.5 meters / 1 second = 5.5 m/s.
Therefore, the average velocity for the time period beginning when t = 2 and lasting 1 second is 5.5 m/s.
To find the average velocity for the time period beginning when t=2 and lasting 1 second, you can use the formula for average velocity:
Average velocity = (change in displacement) / (change in time)
Here, the displacement refers to the change in height, and the change in time is 1 second.
The formula for height in terms of time is given as y = 30t - 4.9t^2.
To find the change in displacement during the 1-second interval, subtract the initial height (at t=2) from the final height (at t=2+1=3).
Initial height (y1) = y(2) = 30(2) - 4.9(2^2) = 60 - 4.9(4) = 60 - 19.6 = 40.4
Final height (y2) = y(3) = 30(3) - 4.9(3^2) = 90 - 4.9(9) = 90 - 44.1 = 45.9
Change in displacement = y2 - y1 = 45.9 - 40.4 = 5.5
Now, divide the change in displacement by the change in time (1 second) to find the average velocity:
Average velocity = change in displacement / change in time = 5.5 m / 1 s = 5.5 m/s
To clarify, the approach you mentioned with the limit definition of derivative is also correct, but in this case, there is a simpler method using the formula for average velocity.