Answer
(a) (i) At t = 1.00 s, $v = -3.80~m/s$.
(ii) At t = 3.00 s, $v = 24.6~m/s$.
(b) The box descends a maximum distance of 4.36 meters below its initial position.
(c) At t = 2.45 seconds, the box returns to its initial position.
Work Step by Step
(a) We can find the acceleration of the box.
$\sum F = ma$
$(36.0~N/s)~t - mg = m~a(t)$
$a(t) = \frac{(36.0~N/s)~t}{3.00~kg} - g$
$a(t) = (12.0~m/s^3)~t - g$
We can use $a(t)$ to find $v(t)$
$v(t) = v_0 + \int_{0}^{t}a(t)~dt$
$v(t) = 0 + \int_{0}^{t}(12.0~m/s^3)~t - g~dt$
$v(t) = (6.00~m/s^3)~t^2 - g~t$
(i) At t = 1.00 s:
$v = (6.00~m/s^3)(1.00~s)^2 - (9.80~m/s^2)(1.00~s)$
$v = -3.80~m/s$
(ii) At t = 3.00 s:
$v = (6.00~m/s^3)(3.00~s)^2 - (9.80~m/s^2)(3.00~s)$
$v = 24.6~m/s$
(b) We can use $v(t)$ to find $x(t)$
$x(t) = x_0 + \int_{0}^{t}v(t)~dt$
$x(t) = 0 + \int_{0}^{t}(6.00~m/s^3)~t^2 - (g~t)~dt$
$x(t) = (2.00~m/s^3)~t^3 - \frac{g}{2}~t^2$
The box reaches its lowest point when $v = 0$. We can find the time $t$ when $v = 0$.
$v = (6.00~m/s^3)~t^2 - g~t = 0$
$t = 0$ or $t = \frac{g}{6.00~m/s^3}$
At $t = \frac{g}{6.00~m/s^3}$:
$x = (2.00~m/s^3)(\frac{g}{6.00~m/s^3})^3 - \frac{g}{2}(\frac{g}{6.00~m/s^3})^2$
$x = -4.36~m$
The box descends a maximum distance of 4.36 meters below its initial position.
(c) $x(t) = (2.00~m/s^3)~t^3 - \frac{g}{2}~t^2 = 0$
$(2.00~m/s^3)~t^3 = \frac{g}{2}~t^2$
$t=0$ or $t = \frac{g}{4.00~m/s^3} = 2.45~s$
At t = 2.45 seconds, the box returns to its initial position.
