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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 meters per second and
t is measured in seconds. What are the units of
int_(1)^(6)f(t)dt ?
seconds
meters
seconds / meter
meters / second
seconds
// meter
^(2)
meters / second
^(2)

The velocity of a bicycle riding down a straight road is measured by the differentiable function $f$ , where $f(t)$ is measured in meters per second and $t$ is measured in seconds. What are the units of $\int_{1}^{6} f(t) d t$ ? $\newline$ seconds $\newline$ meters $\newline$ seconds / meter $\newline$ meters / second $\newline$ seconds $/$ meter $^{2}$ $\newline$ meters / second ${ }^{2}$

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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 meters per second and

t

is measured in seconds. What are the units of

\int_{1}^{6} f(t) d t

?

\newline

seconds

\newline

meters

\newline

seconds / meter

\newline

meters / second

\newline

seconds

/

meter

^{2}

\newline

meters / second

{ }^{2}

Given Information: We are given that $f(t)$ represents the velocity of a bicycle, where $f(t)$ is measured in meters per second ( $\text{m/s}$ ), and $t$ is measured in seconds ( $s$ ). The integral of a velocity function over time gives us the displacement, which is the total distance traveled.
Units Consideration: To find the units of the integral, we can consider the units of the function being integrated (velocity, which is in meters per second) and the units of the variable of integration (time, which is in seconds).
Integration Concept: When we integrate a function with respect to time, we are essentially summing up small products of velocity $\text{(m/s)}$ and time $\text{(s)}$ . The units of these products are meters per second times seconds $\text{(m/s)} \times \text{(s)}$ .
Unit Calculation: Multiplying the units of velocity ( $\text{m/s}$ ) by the units of time ( $\text{s}$ ), we get meters ( $\text{m}$ ) because the seconds ( $\text{s}$ ) in the denominator and the seconds ( $\text{s}$ ) being multiplied cancel out.
Final Result: Therefore, the units of the integral $\int_{1}^{6} f(t) \, dt$ are meters (m), which represent the total displacement or distance traveled by the bicycle over the time interval from $1$ second to $6$ seconds.

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