Final answer:
The integral ∫v(t)dt represents the displacement or distance, the dimension of which is in units of length (L) because it is the product of velocity (LT⁻¹) and time (T).
Step-by-step explanation:
When we interpret the integral ∫v(t)dt, we are essentially finding the area under the velocity-time curve, which represents the displacement or distance traveled, depending on the context and whether or not the motion is one-dimensional. Since velocity (v) is defined as displacement (ds) divided by time (dt), v = ds/dt, its dimensions are length divided by time [LT⁻¹]. Thus, when we integrate velocity over time, the units of ∫v(t)dt are the units of velocity (length/time) multiplied by the units of time, resulting in units of length (L).
The dimension of acceleration (a), which is the derivative of velocity with respect to time, a = dv/dt, is length per time squared [LT⁻²]. Therefore, the units of acceleration times time (adt) are again units of velocity [LT⁻¹]. Finally, the derivatives of acceleration with respect to time (da/dt) would have units of length per time cubed [LT⁻¹T⁻¹], essentially acceleration change rate.