Do I just put in 3 for t?
A ball is dropped from a tower 550 meters above the ground with position function s(t) = 4.9t^2 + 550.
What is the velocity of the ball after 3 seconds?
To find the velocity of the ball after 3 seconds, we need to find the derivative of the position function with respect to time.
The position function is given by s(t) = 4.9t^2 + 550.
The derivative of the position function, denoted as s'(t) or v(t), represents the velocity of the ball at any given time t.
To find the derivative, we apply the power rule, which states that the derivative of t^n is n*t^(n-1).
So, taking the derivative of s(t):
s'(t) = d/dt (4.9t^2 + 550)
= 9.8t
Now, we can substitute t = 3 into the velocity function to find the velocity of the ball after 3 seconds:
v(3) = 9.8 * 3
= 29.4 meters per second
Therefore, the velocity of the ball after 3 seconds is 29.4 meters per second.
To summarize:
1. Calculate the derivative of the position function, s'(t), by applying the power rule.
2. Substitute the given value of t (in this case, t = 3) into the velocity function.
3. Calculate the result to find the velocity of the ball after 3 seconds.