If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.
0.5 seconds
To find the average velocity for a time period, we need to find the change in height and divide it by the change in time.
For the time period beginning when t = 2 and lasting 0.5 seconds, the initial time is t1 = 2 and the final time is t2 = 2 + 0.5 = 2.5.
The initial height is y1 = 48(2) - 16(2^2) = 48(2) - 16(4) = 96 - 64 = 32 ft
The final height is y2 = 48(2.5) - 16(2.5^2) = 48(2.5) - 16(6.25) = 120 - 100 = 20 ft
The change in height is y2 - y1 = 20 - 32 = -12 ft
The change in time is t2 - t1 = 2.5 - 2 = 0.5 seconds
Therefore, the average velocity for the time period of 0.5 seconds starting at t = 2 is:
Average velocity = change in height / change in time = -12 ft / 0.5 seconds = -24 ft/s
estimate the instantaneous velocity when t=2
To estimate the instantaneous velocity at t = 2 seconds, we can calculate the derivative of the height function y = 48t - 16t^2 with respect to t.
The derivative of y with respect to t gives us the rate of change of height with respect to time, which represents the instantaneous velocity.
Taking the derivative of y = 48t - 16t^2, we get:
dy/dt = 48 - 32t
To estimate the instantaneous velocity at t = 2 seconds, we substitute t = 2 into the derivative:
dy/dt = 48 - 32(2) = 48 - 64 = -16 ft/s
Therefore, the estimated instantaneous velocity when t = 2 seconds is -16 ft/s.
To find the average velocity for the time period of 0.5 seconds beginning when t = 2, we need to find the height at t = 2 and t = 2 + 0.5.
Let's start by finding the height at t = 2. Substituting t = 2 into the equation, we get:
y = 48t - 16t^2
y = 48(2) - 16(2)^2
y = 96 - 16(4)
y = 96 - 64
y = 32
So, at t = 2, the height of the ball is 32 feet.
Now, let's find the height at t = 2 + 0.5 = 2.5. Substituting t = 2.5 into the equation, we get:
y = 48t - 16t^2
y = 48(2.5) - 16(2.5)^2
y = 120 - 16(6.25)
y = 120 - 100
y = 20
So, at t = 2.5, the height of the ball is 20 feet.
To find the average velocity, we subtract the initial height from the final height and divide it by the time:
Average velocity = (final height - initial height) / (final time - initial time)
Average velocity = (20 - 32) / (2.5 - 2)
Average velocity = -12 / 0.5
Average velocity = -24 ft/s
Therefore, the average velocity for the time period of 0.5 seconds beginning when t = 2 is -24 ft/s.
To find the average velocity for a specific time period, we need to find the change in the position over that time period and divide it by the duration of the time period.
In this case, we are given the equation for the height of the ball as a function of time, which is given by y = 48t - 16t^2.
To find the change in position (or height) over a time period, we need to find the value of y at the end of the time period minus the value of y at the beginning of the time period.
For the time period beginning when t = 2 and lasting for 0.5 seconds, the starting time is t = 2 seconds and the ending time is t = 2 + 0.5 = 2.5 seconds.
So, we need to find y(2.5) - y(2) to get the change in height over this time period.
Substituting t = 2.5 into the equation y = 48t - 16t^2, we get:
y(2.5) = 48(2.5) - 16(2.5)^2
y(2.5) = 120 - 100
y(2.5) = 20
Similarly, substituting t = 2 into the equation, we get:
y(2) = 48(2) - 16(2)^2
y(2) = 96 - 64
y(2) = 32
Now, we can find the change in height over the time period:
Change in height = y(2.5) - y(2) = 20 - 32 = -12 ft
Finally, to find the average velocity, we divide the change in height by the duration of the time period (0.5 seconds):
Average velocity = Change in height / Duration = -12 ft / 0.5 s = -24 ft/s
So, the average velocity for the time period from t = 2 to t = 2.5 seconds is -24 ft/s.