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The velocity of a bicycle riding down a straight road is measured by the differentiable function
f, where
f(t) is measured in feet per second and
t is measured in seconds. What are the units of
int_(2)^(6)f^(')(t)dt ?
feet
seconds
feet / second
seconds / foot
feet / second
^(2)
seconds / foot
^(2)

The velocity of a bicycle riding down a straight road is measured by the differentiable function $f$ , where $f(t)$ is measured in feet per second and $t$ is measured in seconds. What are the units of $\int_{2}^{6} f^{\prime}(t) d t$ ? $\newline$ feet $\newline$ seconds $\newline$ feet / second $\newline$ seconds / foot $\newline$ feet / second ${ }^{2}$ $\newline$ seconds / foot ${ }^{2}$

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

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

f

, where

f(t)

is measured in feet per second and

t

is measured in seconds. What are the units of

\int_{2}^{6} f^{\prime}(t) d t

?

\newline

feet

\newline

seconds

\newline

feet / second

\newline

seconds / foot

\newline

feet / second

{ }^{2}

\newline

seconds / foot

{ }^{2}

Net Change of Function: The integral of a function's derivative over an interval gives us the net change of the function over that interval. In this case, since $f(t)$ represents the velocity of the bicycle in feet per second, the derivative $f'(t)$ represents the rate of change of velocity, which is acceleration. The units of acceleration would be the units of velocity (feet per second) divided by the units of time (seconds), which gives us feet per second squared (feet/second $^2$ ).
Acceleration and Velocity: When we integrate acceleration over time, we are essentially summing up these small changes in velocity over the time interval. The result of this integral will give us the change in velocity, which has the same units as the original velocity function $f(t)$ , since we are adding up units of acceleration (feet/second $^2$ ) over units of time (seconds). The seconds in the denominator of acceleration will cancel out one of the seconds in the time interval, leaving us with units of feet per second.
Integrating Acceleration: Therefore, the units of the integral from $2$ to $6$ of $f'(t) ext{ dt}$ are the same as the units of the velocity function $f(t)$ , which are feet per second.

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