Answer
(a) $x(t) = 4.00t - \frac{2.00}{3}t^3$
$a_x(t) = -4.00t$
(b) The maximum positive displacement is 3.77 meters.
Work Step by Step
(a) $v_x(t) = 4.00 - 2.00t^2$
$a_x(t) = \frac{dv}{dt} = -4.00t$
$x(t) = \int v_x(t) = \int 4.00 - 2.00t^2$
$x(t) = 4.00t - \frac{2.00}{3}t^3 + c$
Since $x(0) = 0$, then $c=0$.
$x(t) = 4.00t - \frac{2.00}{3}t^3$
(b) The maximum positive displacement occurs when $v_x(t) = 0$ as this is the point where the velocity changes from positive to negative.
$v_x(t) = 0 = 4.00 - 2.00t^2$
$t^2 = 2.00$
$t = \sqrt{2.00}$
We can use this value of $t$ in $x(t)$ to find the maximum positive displacement.
$x(t) = 4.00t - \frac{2.00}{3}t^3$
$x_{max} = (4.00)(\sqrt{2.00}) - \frac{2.00}{3}(\sqrt{2.00})^3$
$x_{max} = 3.77 ~m$
The maximum positive displacement is 3.77 meters.
