# Finding the position $s(t)$ of a particle that has acceleration defined as $a(t)=3t+5$.

I am a Calculus 1 student, and we're learning about antiderivatives. I've run into a problem I'm not sure how to solve.

A particle moves with acceleration defined by $$a(t) = 3t+5$$. Find the position $$s(t)$$ for the particle assuming that $$s(0) = 4$$ and $$v(0) = -1$$.

The question doesn't tell me what $$s$$ and $$v$$ represent and even if I knew what they were, I'm not sure I'd understand the question.

I assume $$s$$ and $$v$$ are functions, but assuming I'm supposed to find $$\int(3t+5)dt$$ at some point, I am not sure if $$s$$ or $$v$$ is $$a'(t)$$ or $$a''(t)$$.

What do I need to understand to solve this problem?

• The velocity $v(t)$ is given by $\int a(t)dt$ and the position is given by $s(t) = \int v(t)dt$. Use the initial conditions to get an exact solution for $v(t)$ and $s(t)$ instead of a family of functions. – nilradical1 Apr 21 at 21:46

Hint :

$$v(t) = \int a(t)\mathrm{d}t$$

$$s(t) = \int v(t) \mathrm{d}t$$

To understand that, think the velocity as the rate of change of the position and the acceleration as the rate of change of the velocity. Thus, they are the first and second derivative of the position respectfully. Thus the inverse way is integration.

So first we want to integrate the acceleration: $$\int{3t+5} \,dt$$

so we get $$v=\frac{3t^2}{2}+5t+c$$, then when $$v(0)=-1$$, so

$$-1=\frac{3*0^2}{2}+(5*0)+c$$, so $$c=-1$$.

Then our expression for $$v(t)=\frac{3t^2}{2}+5t-1$$, so can you do the same thing for speed given that $$s(t)=\int v(t)$$