Exercise 2

\[ \dot{v}_z = \frac{F_z}{m} \]

We integrate this equation on both sides, Leibniz’ notation is very handy here:

\[\int_0^T \frac{\mathrm{d}v_z}{\mathrm{d}t} \mathrm{d}t = \int_0^T \frac{F_z}{m} \mathrm{d}t\]

We have chosen \(T\) for the upper integration limit, since \(t\) is already in use in the integrand. The integral over the derivative of a function is the function itself, so the primitive of the integral on the left side ist just \(v_z\).

\[ \int_{v_z(0)}^{v_z(T)} \mathrm{d}v_z = \frac{F_z}{m} \int_0^T \mathrm{d}t\] \[ \left[ v \right]_{v_z(0)}^{v_z(T)} = \frac{F_z}{m} \left[ t \right]_0^T \] \[ v_z(T) - v_z(0) = \frac{F_z}{m} T \Rightarrow v_z(T) = \frac{F_z}{m} T + v_z(0) \]

\[ v_z(t) = \frac{F_z}{m} t + v_z(0) \]