The position of a bucket falling down a well can be represented by the function h(t)=-16t^{2} where t is time in seconds and h(t) is measured in feet. Based on this function, what is the velocity of the bucket after it has fallen for 4 seconds?
dh/dt is the velocity (1st derivative)
v(t) = -32t
So divide both sides by t and end up with v=-32? That doesn't seem right.
Never mind. I forgot about multiplying it by 4 because it's supposed to be the velocity for 4 seconds.
I am shocked to see your first try.
v(t) is a function.
it is v of t, not v times t!
To find the velocity of the bucket after it has fallen for 4 seconds, we need to find the derivative of the function h(t) with respect to t. The derivative will give us the rate of change of the position, which is the velocity.
The given function represents the height of the bucket as a function of time, h(t) = -16t^2.
To find the derivative, we apply the power rule for derivatives. The power rule states that if we have a function f(t) = c*t^n, where c is a constant and n is any real number, the derivative of this function is f'(t) = c*n*t^(n-1).
Applying the power rule to the given function, we have h'(t) = (-16)*2*t^(2-1) = -32t.
So, the derivative of h(t) is h'(t) = -32t.
Now, we can find the velocity of the bucket after it has fallen for 4 seconds by substituting t = 4 into the derivative function:
h'(4) = -32(4) = -128 feet per second.
Therefore, the velocity of the bucket after it has fallen for 4 seconds is -128 feet per second. Note that the negative sign indicates that the bucket is moving downwards.