Acceleration of a particle is given as $a = 0.1(t-5)^2$. The particle is initially at rest.
Here is the way I initially attempted it. (I didn't use substitution as the question in the textbook appears before substitution is officially taught.)
$$a\:=\:0.1t^2-t+2.5$$ so $$v = \frac{1}{30}t^3-0.5t^2+2.5t+c$$ $c$ evaluates to zero.
But my textbook uses a substitution and gets:
$$\frac{1}{30}\left(t-5\right)^3 +c$$
In this case, c has a value when t= 0 and so the end result is different. Wouldn't this lead to different values in general?
Note that you ended up with $$ v(t) = \frac{1}{30} t^3 - \frac12 t^2 + \frac52 t $$
But the book's answer must have $c = 5^3/30$, so the complete answer is $$ \begin{split} v(t) &= \frac{1}{30} (t-5)^3 + \frac{5^3}{30}\\ &= \frac{(t-5)^3-5^3}{30} \\ &= \frac{t^3 - 3\cdot 5 t^2 + 3 \cdot 5^2t - 5^3 + 5^3}{30} \\ &= \frac{t^3}{30} - \frac{t^2}{2} + \frac{5t}{2}\\ \end{split} $$ which is the same answer as the one you ended up with.