Acceleration to be in m/s^2 and integrate w.r.t. x

[negatifzeo]

I'm not sure if this is the write forum for this question, but I'm sure someone(maybe everyone) here knows the answer to my question. My question has to do with units and integration/differentiation. Does something "happen" to these units during these operations, or are you just supposed to "know"? For example, when you take acceleration to be in m/s^2 and integrate w.r.t. x, can it be "shown in the work" that the units will be m/s?

 

[Pengwuino]

I assume you mean integrating with respect to time. When you integrate, the dt gives you the [time] dimension that gives you m/s as the final result. Similarly, when you differentiate with respect to, for example, time, you're attaching $$\frac{d}{dt}$$ your equation which gives [1/time] dimensions.

 

[elibj123]

Integrating involves multiplication, while summation doesn't change the units (off course it's important that you are summing sizes with the same units). The dt is not there just for notation.

 

[HallsofIvy]

As both pengwuino and elibj123 said, integration is essentially like multiplication and differentiation like division. If you are integrating with respect to a variable having particular units, the units of the result are the units of the integrand times the units of the variable. If you are differentiating a function with respect to a variable, the units of the derivative are the units of the function being differentiated divide by the units of the variable.

If you differentiate distance, x, in units of meters, with respect to time, t, in seconds, the result is a speed, dx/dt, with units of m/s, meters per second.

If you have a have an object with density, $\rho$ in units of $g/m^3$, grams per cubic meter, and integrate over its volume with respect ot x, y, and z in units of meters, then the mass, \(\displaystyle \int\int\int \rho(x,y,z)dxdydz\), has units of $(g/m^3)(m)(m)(m)= g$, grams.