The convective term in Reynolds Transport theorem

Question

Reynolds Transport theorem states that $$\frac{D B_{sys}}{Dt}=\frac{\partial B_{CV}}{\partial t}-\dot{B_{in}}+\dot{B_{out}}$$ where $B_{in,t+\Delta t}=b_1m_{1,t+\Delta t}=b_1\rho_{1}\forall_{in,t+\Delta t}=b_1\rho_1V_1\Delta tA_1$, then

$$\dot{B_{in}}=\lim_{\Delta t\to0}\frac{B_{in,t+\Delta}}{\Delta t}=\lim_{\Delta t\to0}\frac{b_1\rho_1V_1\Delta tA_1}{\Delta t}=b_1\rho_1V_1A_1$$

Now my question is, how come volume $\forall_{in}$ at $t+\Delta t$ is equal to $V_1 \Delta tA_1$?, where $V_1$ is the velocity and $A_1$ is the area. Dimension-ally, this relation works but still do not understand the physics.

Answer 1

$V_1 A_1$ is basically the volumetric flowrate. So in a time $\Delta t$, a particle moving with flowrate $V_1 A_1$, sweeps out a volume $V_1 A_1 \Delta t$.