I don't understand this at all!
use the position equation s=-16t^2 +v0t+s0 to write a function that represents the situation. Find the average rate of change from t1 to t2.
a) An object is thrown upward from a height of 6 feet at a velocity of 64 feet per second.... t1=0 , t2=3
b) An object is thrown upward from ground level at a velocity of 120 feet per second.... t1=3, t2=5
Well, aren't you in for a fun math ride! Let's break this down with a tinge of humor, shall we?
a) For the first situation, we have an object being thrown upward from 6 feet with a velocity of 64 feet per second. Now, let's plug in the values into the position equation and see what we get. We have s = -16t^2 + v0t + s0. Putting the given values in, we get s = -16t^2 + 64t + 6.
Now, to find the average rate of change from t1 = 0 to t2 = 3, we'll put these values into the function. So, brace yourself for some number crunching and remember, it's all just math chuckles. Calculate s(3) - s(0) and divide it by 3 - 0. You'll get your answer!
b) In the second situation, we have an object being thrown upward from ground level with a velocity of 120 feet per second. Following the same process, we plug in the values into the position equation and get s = -16t^2 + 120t.
To find the average rate of change from t1 = 3 to t2 = 5, we'll use the same method. Calculate s(5) - s(3) and divide it by 5 - 3. The answer will reveal itself!
Remember, even though math can be confusing at times, don't forget to laugh along the way. Good luck with your calculations!
To write a function that represents the situation using the position equation s = -16t^2 + v0t + s0, we need to substitute the given values of v0 and s0 into the equation.
a) For the first scenario where an object is thrown upward from a height of 6 feet at a velocity of 64 feet per second, we have v0 = 64 and s0 = 6. So the function representing this situation is:
s(t) = -16t^2 + 64t + 6
b) For the second scenario where an object is thrown upward from ground level at a velocity of 120 feet per second, we have v0 = 120 and s0 = 0 (since it starts from ground level). So the function representing this situation is:
s(t) = -16t^2 + 120t + 0
Now, to find the average rate of change from t1 to t2, we need to calculate the difference in the function values at t1 and t2, and then divide by the difference in the time values (t2 - t1).
a) For t1 = 0 and t2 = 3 in the first scenario:
Average rate of change = (s(3) - s(0))/(3 - 0)
= (-(16 * 3^2) + 64 * 3 + 6 - (-(16 * 0^2) + 64 * 0 + 6))/3
= (-(16 * 9) + 192 + 6 - 0)/3
= (-144 + 192 + 6)/3
= 54 feet per second
b) For t1 = 3 and t2 = 5 in the second scenario:
Average rate of change = (s(5) - s(3))/(5 - 3)
= (-(16 * 5^2) + 120 * 5 + 0 - (-(16 * 3^2) + 120 * 3 + 0))/(5 - 3)
= (-(16 * 25) + 600 - (-(16 * 9) + 120 * 3))/2
= (-400 + 600 - (-144 + 360))/2
= (200 - 216)/2
= -8 feet per second
Therefore, the average rate of change from t1 to t2 in the first scenario is 54 feet per second, and in the second scenario is -8 feet per second.
To understand the problem and find the average rate of change, we need to use the given position equation:
s = -16t^2 + v0t + s0
In this equation, s represents the position of the object at a given time t, -16t^2 is the gravitational term, v0t represents the initial velocity multiplied by time, and s0 represents the initial position.
To find the function that represents the situation, we need to substitute the given values into the position equation.
a) For the first situation, an object is thrown upward from a height of 6 feet at a velocity of 64 feet per second. Here, the initial position s0 is 6 feet and the initial velocity v0 is 64 feet per second. Since the object is thrown upward, the initial velocity is positive. Therefore, the function that represents this situation is:
s(t) = -16t^2 + 64t + 6
b) For the second situation, an object is thrown upward from ground level, so the initial position s0 is 0 feet. The initial velocity v0 is 120 feet per second, and since the object is thrown upward, the initial velocity is positive. The function representing this situation is:
s(t) = -16t^2 + 120t
To find the average rate of change from t1 to t2, we need to calculate the change in position divided by the change in time.
In both scenarios, t1 and t2 are given. Let's calculate the average rate of change for each case:
a) For the first scenario, where t1 = 0 and t2 = 3:
Average rate of change = [s(t2) - s(t1)] / [t2 - t1]
= [s(3) - s(0)] / [3 - 0]
= [(-16 * 3^2) + (64 * 3) + 6] / [3 - 0]
= [-144 + 192 + 6] / 3
= 54 / 3
= 18 feet per second
b) For the second scenario, where t1 = 3 and t2 = 5:
Average rate of change = [s(t2) - s(t1)] / [t2 - t1]
= [s(5) - s(3)] / [5 - 3]
= [(-16 * 5^2) + (120 * 5)] - [(-16 * 3^2) + (120 * 3)] / [5 - 3]
= [-400 + 600] - [-144 + 360] / [2]
= 200 - 216 / 2
= -16 / 2
= -8 feet per second
Thus, the average rate of change in scenario a) is 18 feet per second, and in scenario b) it is -8 feet per second.