Please help It due today MATH!!1
2 answers:
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Check the picture below.
now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.
now, those are the values when the rocket is 30 feet up above.
now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.
now, those are the values when the rocket is 30 feet up above.
Answer:
C
Step-by-step explanation:
Since we are given the position function, the average velocity will simply be the average slope of the position function.
We want to find the average velocity for time 0≤t≤3.
So, our endpoints are t = 0 and t = 3.
By substitution, the position of the particle at those times are:
So, we have the two points (0, 0) and (3, -45).
Then by the slope formula, we acquire that:
So, the average velocity of the particle is about -15.
This implies that the particle was mainly moving backwards for time 0≤t≤3.
Hence, our final answer is C.
*If you differentiated first and then did the steps above, then you found the average acceleration of the particle (in this case it stays constant), not the average velocity. Remember that the slope of the position graph is velocity, and the slope of the velocity graph is acceleration!