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The position of a bicycle riding down a straight road is measured by the differentiable function
f, where
f(t) is measured in feet and
t is measured in minutes. What are the units of
f^('')(t) ?
feet
minutes
feet / minute
minutes / foot
feet / minute
^(2)
minutes / foot
^(2)

The position of a bicycle riding down a straight road is measured by the differentiable function $f$ , where $f(t)$ is measured in feet and $t$ is measured in minutes. What are the units of $f^{\prime \prime}(t)$ ? $\newline$ feet $\newline$ minutes $\newline$ feet / minute $\newline$ minutes / foot $\newline$ feet / minute ${ }^{2}$ $\newline$ minutes / foot ${ }^{2}$

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. The position of a bicycle riding down a straight road is measured by the differentiable function

f

, where

f(t)

is measured in feet and

t

is measured in minutes. What are the units of

f^{\prime \prime}(t)

?

\newline

feet

\newline

minutes

\newline

feet / minute

\newline

minutes / foot

\newline

feet / minute

{ }^{2}

\newline

minutes / foot

{ }^{2}

Understand function and question: Understand the given function and what is being asked. $\newline$ The function $f(t)$ represents the position of a bicycle along a straight road, where $f$ is measured in feet and $t$ is measured in minutes. We are asked to find the units of $f''(t)$ , which is the second derivative of the position function with respect to time.
Determine units of first derivative: Determine the units of the first derivative $f'(t)$ . The first derivative $f'(t)$ represents the velocity of the bicycle. Since velocity is the rate of change of position with respect to time, the units of $f'(t)$ will be the units of $f$ divided by the units of $t$ , which is feet per minute (feet/minute).
Determine units of second derivative: Determine the units of the second derivative $f''(t)$ . The second derivative $f''(t)$ represents the acceleration of the bicycle. Acceleration is the rate of change of velocity with respect to time. Since we have already determined that the units of velocity ( $f'(t)$ ) are feet per minute, the units of acceleration will be the units of velocity divided by the units of time. Therefore, the units of $f''(t)$ will be (feet/minute) per minute, which simplifies to feet per minute squared (feet/minute $^2$ ).

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