How Do I Integrate Kinematics Problems with Units?

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    Integrating Units
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Homework Help Overview

The discussion revolves around integrating kinematics problems, specifically focusing on the relationship between velocity and displacement, using the example of the equation velocity = Be^(-rt). Participants are exploring the units involved in the integration process.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how the integral of velocity results in displacement and questions the units involved in the integration process. Some participants clarify the role of the variable "t" in the equation and its impact on the dimensional analysis.

Discussion Status

Participants are actively engaging with the problem, providing clarifications regarding the units of time and the dimensionless nature of the exponential term. There appears to be a productive exchange of ideas, with some participants confirming their understanding of the concepts discussed.

Contextual Notes

There is a focus on ensuring that all variables are correctly accounted for in terms of their units, particularly the importance of "t" being in seconds to maintain dimensional consistency in the integration process.

noypingpogi
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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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