What's the difference between $v = a\cdot t$ and $\vec{v} = \int \vec{a} \, \mathrm dt$

Question

In highschool, I learned $v = at$ and in university, I am learning $\vec{v} = \int \frac{\vec{F}}{m} \, \mathrm dt = \int \vec{a} \, \mathrm dt$.

I understand one is for $v= at$ is for one-dimension and the latter for multiple dimensions. However, I don't understand why in one dimension, we don't do $v = \int a(t) \, \mathrm dt$ but rather multiply it by the time to get the acceleration at time $t$. Shouldn't the acceleration accumulate and therefore do the integral instead?

I am confused.

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As an example,

A particle of mass $m=2$ is acted on by a force $$ \mathbf{F}=\left(4 t, 6 t^{2},-4 t\right) $$ At $t=0,$ the particle has velocity zero and is located at the point $(1,2,3)$ .
Find the velocity vector $\mathbf{v}(t)$ for $t \geq 0$

We can easily know that $\vec{a} = \langle 2t,3t^2,-2t\rangle$. However, the velocity is not $\vec{a}\cdot t$ (which is possible with no problem since $t$ is a scalar and it still returns a vector), but rather anti-integral of the vector?

Answer 1

The formula $v = v_0 + at$ assumes that the acceleration is constant. The formula $v = v_0 + \int_{t_0}^{t_f} a(t) \, dt$ allows for the possibility that the acceleration changes with time.

Answer 2

The first formula $v=at$ is valid for a constant whereas the second one is valid also for not constant acceleration that is $v=\int_0^t a(u)du$.