How Do I Integrate Kinematics Problems with Units?

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    Integrating Units
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Integrating kinematics problems requires careful attention to units, particularly when dealing with exponential functions. In the example of velocity given as Be^(-rt), where B is in m/s and r is in s^-1, the integration with respect to time (t) results in displacement measured in meters. The integration process shows that the units of velocity (m/s) multiplied by time (s) yield meters, confirming the displacement. The exponential term e^(-rt) is dimensionless, as the product of r and t cancels out the units. Understanding these unit conversions is crucial for correctly solving kinematics problems.
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hello, I'm new to the forums. Can someone help me with integrating kinematics problems? For example velocity= Be^(-rt), where B= 3.00 m/s and r=0.500 s^-1. i don't understand how the integral's unit becomes m (since the integral of velocity is displacement). someone help me! thanks
 
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noypingpogi said:
hello, I'm new to the forums. Can someone help me with integrating kinematics problems? For example velocity= Be^(-rt), where B= 3.00 m/s and r=0.500 s^-1. i don't understand how the integral's unit becomes m (since the integral of velocity is displacement). someone help me! thanks
You forgot the unit of "t". "rt" is a dimensionless number. Velocity has the unit of "m/s", and if you integrate in respect to "t" you have to multiply this unit by "s".
 
\int B e^{rt} \mathrm{d}t = \frac{B}{r} e^{rt}
On the left, you have B (m/s) times dt (s), giving units of meters; on the right, B (m/s) divided by r (1/s), again giving meters. As A.T. said, the exponential part is dimensionless.
 
yes, i figured it out. "t" is in seconds which makes e^-rt dimensionless (since s^-1 * s = 1). thanks for the help.
 
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