As a particle moves along the number line, its position at time $t$ is $s(t)$, its velocity is $v(t)$, and its acceleration is $a(t)=3 t^{2}$.

If $v(0)=3$ and $s(0)=1$, what is $s(2)$ ?
$$
s(2)=
$$
$\square$

Question

As a particle moves along the number line, its position at time $t$ is $s(t)$, its velocity is $v(t)$, and its acceleration is $a(t)=3 t^{2}$.

If $v(0)=3$ and $s(0)=1$, what is $s(2)$ ?
$$
s(2)=
$$
$\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $s(2)$, we first need to find the velocity function $v(t)$ by integrating the acceleration function $a(t) = 3t^2$. ‖ ‖ step 2 ⋮ Using the asksia-ll calculation list, the integral of $3t^2$ from 0 to 2 is 8. This gives us the change in velocity from $t=0$ to $t=2$. ‖ ‖ step 3 ⋮ Since $v(0) = 3$, we add this initial velocity to the change in velocity to find $v(2)$. So, $v(2) = 3 + 8 = 11$. ‖ ‖ step 4 ⋮ Next, we integrate the velocity function to find the position function $s(t)$. The integral of a constant 3 from 0 to 2 is 6, which represents the change in position due to the initial velocity over the time interval. ‖ ‖ step 5 ⋮ We add the initial position $s(0) = 1$ to the change in position to find $s(2)$. So, $s(2) = 1 + (8) + (6) = 15$. ‖ ∻Answer∻ ⚹ $s(2) = 15$ ⚹ ∻Key Concept∻ ⚹Integration of acceleration to find velocity, and integration of velocity to find position⚹ ∻Explanation∻ ⚹The position $s(2)$ is found by integrating the acceleration function to get the velocity, then integrating the velocity to get the position, and considering the initial conditions for velocity and position.⚹◊What is the formula for the relationship between position, velocity, and acceleration of a particle in one dimension?⍭ Generate me a similar question◊

Sia