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) = 4.9t2 + 550.
What is the velocity of the ball after 3 seconds?
supposed to be a 4.9t^2+550
no, s(t) is the position function. You want the velocity, which is
ds/dt = -9.8t
then you can sub in t=3.
And, since the ball is falling,
s(t) = -4.9t^2 + 550
so would the velocity be -29.4?
To find the velocity of the ball after 3 seconds, you need to calculate the derivative of the position function s(t) with respect to time.
The velocity function v(t) is the derivative of the position function s(t). In this case, s(t) is given as 4.9t^2 + 550, so we can find v(t) by taking the derivative of s(t) with respect to t.
To do this, take the derivative of each term separately. The derivative of 4.9t^2 is (2)(4.9t) = 9.8t, since the derivative of t^n is n*t^(n-1). The derivative of 550 is 0, since it is a constant.
Therefore, the velocity function v(t) = 9.8t.
Now, to find the velocity of the ball after 3 seconds, you can substitute t = 3 into the velocity function v(t):
v(3) = 9.8 * 3
= <<9.8*3=29.4>>29.4
So, the velocity of the ball after 3 seconds is 29.4 meters per second (m/s).
In summary, to find the velocity of the ball after 3 seconds:
1. Calculate the derivative of the position function s(t) with respect to t.
2. Substitute t = 3 into the resulting velocity function v(t) to find the velocity after 3 seconds.