If the position function in feet for a free-falling particle on Planet X is s(t)= -8t^2 +16 + 64, what is the velocity of the after 2 seconds in feet per second?
A. -32
B. -16
C. 16
D. 64
Find the derivative of the function to the velocity function, and plug in 2 for t.
How do I find the derivative?
Use the power rule:
https://www.mathsisfun.com/calculus/derivatives-rules.html
To find the velocity of the particle after 2 seconds, we need to differentiate the position function with respect to time. The derivative of the position function will give us the velocity function.
Given that the position function is s(t) = -8t^2 + 16t + 64, we can differentiate it using the power rule of derivatives.
First, differentiate each term separately:
d/dt (-8t^2) = -16t
d/dt (16t) = 16
d/dt (64) = 0 (since it's a constant)
Combine the derivatives:
v(t) = -16t + 16 + 0
v(t) = -16t + 16
Now, substitute t = 2 into the velocity function to find the velocity after 2 seconds:
v(2) = -16(2) + 16
v(2) = -32 + 16
v(2) = -16
Therefore, the velocity of the particle after 2 seconds is -16 feet per second.
So the correct answer is B. -16.