Kinematics equations with calculus what does dv mean?

Question

I am trying to solve for kinematics equations using calculus. I am trying to get to this equation:

$$v_t = v_0 + at$$

Which comes from,

$$a = \frac{\mathrm{d}v}{\mathrm{d}t}$$ $$ \mathrm{d}v = a \cdot \mathrm{d}t$$ $$ \int_{v_0}^{v_t} \mathrm{d}v = \int_0^{\Delta t} a \cdot \mathrm{d}t$$ $$ v_t - v_0 = a \cdot t$$ $$ \boxed{ v_t = v_0 + at}$$

1) What does this mean? Does it mean that the change in velocity = the acceleration times the change of time?

Now, when you continue, I also understand the definite integral. I understand why you must have v and v initial and t and 0 (t initial). However, my second question is:

2) How can you get the antiderivative of something that is not even a derivative? is dv a derivative? I thought the derivative was (dv/dt), not just dv by itself?

I just don't understand how dv can be anything by itself?

Answer 1

Answer (1):

Yes, That is what it means.

$v-v_o$(change in velocity)=$a(t-t_o)$(change in time)

In your case $t_o$ is $0$.

Answer (2):

dv represents a very little change in velocity, i.e. change in velocity when change in time$\approx0$

Answer 2

1) Yes, it means the change in velocity equals the acceleration times the change in time.

2) $\int\mathrm dv$ can also be written as $\int 1\,\mathrm dv$. Then it becomes more clear that the antiderivative is $v$, since the "derivative" is $1$. That's not the only way to think of it though, so there's no point in writing the extra $1$.