The velocity of a ball that has been tossed vertically in the air is given by v(t) = 16 − 32t, where v is measured in feet per second, and t is measured in seconds. The ball is in the air from t = 0 until t = 2.
Determine the value of v'(1).
What are the units on the value of v'(1)?
Thanks!
v = ft/s
dv/dt is thus ft/s^2 (acceleration)
Since gravity is constant, v' is constant.
since v'(t) = a(t), which is the acceleration
a(t) = -32
since a(t) is a constant, we get the acceleration due to gravity at -32 ft/s^2
and v'(1) = a(1) = -32 ft/s^2
To determine the value of v'(1), we need to find the derivative of the function v(t) = 16 - 32t and evaluate it at t = 1.
First, let's find the derivative of v(t). The derivative of a function describes its rate of change with respect to the independent variable. In this case, we're looking for the rate of change of velocity with respect to time.
v(t) = 16 - 32t
To find the derivative, we can apply the power rule and constant rule of differentiation.
The power rule states that if we have a function of the form f(t) = kt^n, where k is a constant and n is any real number, then the derivative of f(t) with respect to t is given by f'(t) = nkt^(n-1).
In our case, we have v(t) = 16 - 32t, where n = 1 and k = -32. Therefore, applying the power rule, we get:
v'(t) = -32 * 1 * t^(1-1) = -32
So, the derivative of v(t) with respect to t is v'(t) = -32.
Now, we can evaluate v'(1) by substituting t = 1 into the derivative:
v'(1) = -32
Therefore, the value of v'(1) is -32 feet per second.
Next, let's determine the units on the value of v'(1). The original velocity function v(t) was given in feet per second, and we determined that the derivative v'(t) is also in feet per second. Therefore, the units on the value of v'(1) are feet per second.
In summary:
- The value of v'(1) is -32 feet per second.
- The units on the value of v'(1) are feet per second.