the height in feet of a rocket from ground level is given by the function f(t)= -16t^2+160t. what is the instantaneous velocity of the rocket 3 seconds after it is launched?

A 32 feet/sec
B 64 feet/sec
C 12 feet/sec
D 10 feet/sec
E 8 feet/sec

okay - another power example. What do you get?

do i plug 3 in for t?

first take the derivative, which is the velocity at any instant.

ds/dt = ft/s, or velocity

Plug 3 into that.

its going to be 32 hun

Hmm. How did you get that, jessica?

f'(t) = -32t + 160
f'(3) = -96 + 160 = 64

To find the instantaneous velocity of the rocket at a given time, we need to take the derivative of the height function f(t) with respect to time. In this case, the height function is given by f(t) = -16t^2 + 160t.

Taking the derivative of f(t) with respect to time t, we get:

f'(t) = d/dt(-16t^2 + 160t)

To find the derivative, we can apply the power rule of differentiation. The power rule states that for a function of the form h(t) = at^n, the derivative is given by h'(t) = nat^(n-1). Applying this rule:

f'(t) = -16 * 2t^(2-1) + 160 * 1t^(1-1)
= -32t + 160

Now that we have the derivative, we can find the instantaneous velocity 3 seconds after the rocket is launched by plugging in t = 3 into f'(t):

f'(3) = -32(3) + 160
= -96 + 160
= 64 feet/sec

Therefore, the instantaneous velocity of the rocket 3 seconds after it is launched is 64 feet/sec.
The correct answer is option B: 64 feet/sec.