If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet t seconds later is given by y = 40t − 16t^2.
Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.
1) 0.5 seconds.
when t = 2, y = 40(2) - 16(4) = 16
when t = 2.5 , y = 40(2.5) - 16(2.5^2) = 0
avg velocity = (0-16)/(2.5-2) = -32 ft/s
-32
To find the average velocity for a given time period, we need to calculate the change in position (or height) over that time period divided by the duration of the time period.
Given that the height at time t is given by y = 40t - 16t^2, we can find the change in position over the time period beginning at t = 2 and lasting for 0.5 seconds.
Let's denote the initial time as t1 = 2 and the final time as t2 = 2 + 0.5 = 2.5.
To find the change in position, we need to evaluate y(t2) - y(t1).
To find the average velocity for a given time period, we need to calculate the displacement of the object during that time period and divide it by the duration of the time period.
In this case, the height of the ball is given by the equation y = 40t - 16t^2. To find the displacement between t = 2 and t = 2.5, we need to find the value of y at each of these times and subtract them.
First, let's find the height at t = 2. Substituting t = 2 into the equation, we get:
y = 40 * 2 - 16 * 2^2
y = 80 - 64
y = 16 ft
Next, let's find the height at t = 2.5:
y = 40 * 2.5 - 16 * 2.5^2
y = 100 - 16 * 6.25
y = 100 - 100
y = 0 ft
To find the displacement, we subtract the height at t = 2 from the height at t = 2.5:
Displacement = 0 ft - 16 ft
Displacement = -16 ft
Now, we divide the displacement by the duration of the time period, which is 0.5 seconds:
Average velocity = Displacement / Time
Average velocity = -16 ft / 0.5 s
Average velocity = -32 ft/s
Therefore, the average velocity for the time period of 0.5 seconds, starting from t = 2, is -32 ft/s.