# Integrating velocity, how to get to this given equation?

Assuming every parameter is constant except for the variable $$t$$, how does the author get from

to here ?

When I integrate velocity myself I get $$z(t)=v_0t+\frac{v_e(m_0-qt)(\ln(1-\frac{qt}{m_0})-1)}{q} -\frac{gt^2}{2}$$ which leads to totally different results when using the same parametric and variable values. Specifically I can't understand how does he get rid of the $$-1$$ in the second parenthesis, and where does $$(v_0+v_e)t$$ come from.

There is an implicit assumption that you may have overlooked here.

In physics, the convention is that the initial position is zero, i.e. $$z(0) = 0$$. Therefore, you should first write your integral as,

$$z(t) = Z + v_0t+\frac{v_e(m_0-qt)}{q} \left[ \ln\left(1-\frac{qt}{m_0}\right)-1 \right]-\frac{gt^2}{2} \tag{1}$$

Then, find out $$Z$$ by setting $$z(0)=0$$, which yields,

$$Z = \frac{v_e m_0}{q}$$

Now, plug the value of $$Z$$ back into (1), you get

$$z(t) = (v_0+v_e)t+\frac{v_e(m_0-qt)}{q}\ln \left(1-\frac{qt}{m_0} \right) -\frac{gt^2}{2}$$

The resulting $$z(t)$$ is guaranteed to be zero at $$t = 0$$.

• I forgot to take the integration constant into account, mainly because I was thinking at it as a definite integral $\int_0^t'v(t)dt$. Thank you. :) Aug 17, 2019 at 8:53

\begin{align}\frac{v_e\Big(m_0-qt\Big)\Big(\ln(1-\frac{qt}{m_0})-1\Big)}{q}&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)-1\Big)\\&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)\Big)-v_e\Big(\frac{m_0}{q}-t\Big)\\&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)\Big)+v_et-v_e\Big(\frac{m_0}{q}\Big)\end{align}
differs from $$z(t)$$ by the constant value of $$v_e\Big(\frac{m_0}{q}\Big)$$. This is due to the integration constant found by setting $$t=0$$ and then solving $$z(0)=0$$.
Hint: $$\int ln \, (1-at)dt=-\frac 1 a\int ln \, s \, ds$$ where $$s=1-at$$ and $$\int ln \, s \, ds=sln\, s -s+C$$.
See my comment below for determining $$C$$.
• Your answer defers from the given answer only by a constant. Remember that when you integrate you have to add an arbitrary constant. The exact value of the constant can be determined by putting $t=0$: the initial velocity is $v_0$. Aug 17, 2019 at 5:22