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Suppose we have a ball traveling with an acceleration given by the function $$a(t) = (t-t_0)$$ and initial velocity $v_0$ at time $t_0$, then we can integrate $a(t)$ to obtain the equation for velocity with time $v(t)$.

The problem I have is, integrating the function $a(t)$ different ways seemed to yield different results, but I can't find where I went wrong. Please take a look at the integrations I did below. Now $v_0$ should be replacing the constants $c$ but the two equations are obviously not equal. Is it because in mathematics, the term ${t_0}^2 \over 2$ is actually part of the constant $c$? Is one of the equations wrong and the other correct? If so, then why?

$$\int (t-t_0) \ dt = \frac {1}{2}(t-t_0)^2 + c \\ \int (t-t_0) \ dt = \frac {t^2}{2} -tt_0 + c $$

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    $\begingroup$ Yes, the two constants of integration differ by a factor $\frac 12 t^2_0$. which is a constant $\endgroup$ – Farcher Feb 18 '18 at 16:03
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Yes - in the second equation, you have absorbed the constant $\frac{t_0^2}{2}$ into the constant $c$. Remember that $c$ is arbitrary, and only takes on a meaningful value when you impose initial conditions on your solution. If you set $v(t_0) = v_0$, then you will find that your solutions agree with one another - though the actual value that $c$ takes on will be different in each case.

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  • $\begingroup$ Thank you very much. Now I see, first equation $c = v_0$ and second equation $c$ is different. I had the same idea from mathematics that $c$ is normally arbitrary, but I became unsure since one of the two equations I entered showed 'incorrect' in an online assignment. $\endgroup$ – user147526 Feb 18 '18 at 16:26

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