Question

An object has a constant acceleration of 30 ft/sec2, an initial velocity of -10 ft/sec, and an initial position of 4 ft. Find the position function, s(t), describing the motion of the object.

Answer 1

Let $a(t),v(t),s(t)$ be the acceleration, velocity, and position functions, respectively. Then

$v(t)=s'(t)$
$a(t)=v'(t)=s''(t)$

Since the acceleration is a constant 30 ft/s^2, you have

$a(t)=s''(t)=30$
$\displaystyle\int s''(t)\,\mathrm dt=\int30\,\mathrm dt$
$\displaystyle s'(t)=v(t)=30t+C_1$

You're told that the initial velocity is $v(0)=-10$, so you get

$-10=30(0)+C_1\implies C_1=-10$

Now $s'(t)=30t-10$. Integrating once more, you find

$\displaystyle\int s'(t)\,\mathrm dt=\int(30t-10)\,\mathrm dt$
$s(t)=15t^2-10t+C_2$

and since $s(0)=4$, you get

$4=15(0)^2-10(0)+C_2\implies C_2=4$

which means the exact position function is

$s(t)=15t^2-10t+4$