Answer
(a) The box moves a distance of 14.0 meters to the right.
(b) The speed is 18.0 m/s at t = 3.00 s.
Work Step by Step
$F(t) = (6.00~N/s^2)~t^2$
$a(t) = \frac{F(t)}{m} = \frac{(6.00~N/s^2)~t^2}{2.00~kg}$
$a(t) = (3.00~m/s^4)~t^2$
Note that we can let $a(t) = -(3.00~m/s^4)~t^2$ since the force is directed to the left.
We can use $a(t)$ to find $v(t)$.
$v(t) = v_{0x}+ \int_{0}^{t}a_x(t)~dt$
$v(t) = 9.00~m/s+ \int_{0}^{t} -(3.00~m/s^4)~t^2~dt$
$v(t) = 9.00~m/s- (1.00~m/s^4)~t^3$
We can use $v(t)$ to find $x(t)$.
$x(t) = x_0+ \int_{0}^{t}v(t)~dt$
$x(t) = 0+ \int_{0}^{t}9.00~m/s- (1.00~m/s^4)~t^3~dt$
$x(t) = (9.00~m/s)~t- (0.25~m/s^4)~t^4$
We can find the time $t$ when $v=0$.
$v = 9.00~m/s- (1.00~m/s^4)~t^3 = 0$
$t^3 = 9.00~s^3$
$t = 2.08~s$
We can find $x$ when $t = 2.08~s$.
$x = (9.00~m/s)(2.08~s)- (0.25~m/s^4)(2.08~s)^4$
$x = 14.0~m$
The box moves a distance of 14.0 meters to the right.
(b) We can find $v$ when t = 3.00 s.
$v = 9.00~m/s- (1.00~m/s^4)~t^3$
$v = 9.00~m/s- (1.00~m/s^4)(3.00~s)^3$
$v = -18.0~m/s$
The speed is 18.0 m/s at t = 3.00 s.
