Answer
(a) When $v_{0x} = -4.00~m/s$, the particle has the same x-coordinate at t = 0 and t = 4.00 s.
(b) When t = 4.00 s, the velocity will be 12.0 m/s
Work Step by Step
$a_x(t) = -2.00~m/s^2 + (3.00~m/s^3)~t$
(a) We can use $a_x(t)$ to find $v_x(t)$.
$v_x(t) = v_{0x}+ \int_{0}^{t}a_x(t)~dt$
$v_x(t) = v_{0x}+ \int_{0}^{t}-2.00~m/s^2 + (3.00~m/s^3)~t~dt$
$v_x(t) = v_{0x}+ (-2.00~m/s^2)~t + (1.50~m/s^3)~t^2$
We can use $v_x(t)$ to find $x(t)$.
$x(t) = x_0+ \int_{0}^{t}v_x(t)~dt$
$x(t) = x_0+ \int_{0}^{t}v_{0x}+ (-2.00~m/s^2)~t + (1.50~m/s^3)~t^2~dt$
$x(t) = x_0 + v_{0x}~t+ (-1.00~m/s^2)~t^2 + (0.50~m/s^3)~t^3$
When t = 0,
$x = x_0 + v_{0x}(0)+ (-1.00~m/s^2)(0)^2 + (0.50~m/s^3)(0)^3$
$x = x_0$
When t = 4.00 s,
$x = x_0 + v_{0x}(4.00~s)+ (-1.00~m/s^2)(4.00~s)^2 + (0.50~m/s^3)(4.00~s)^3$
$x = x_0 + v_{0x}(4.00~s) -16.0~m+32.0~m$
We need these two x-coordinates to be equal.
$x_0 + v_{0x}(4.00~s) -16.0~m+32.0~m = x_0$
$v_{0x} = \frac{-16.0~m}{4.00~s} = -4.00~m/s$
When $v_{0x} = -4.00~m/s$, the particle has the same x-coordinate at t = 0 and t = 4.00 s.
(b) When t = 4.00 s,
$v = (-4.00~m/s)+ (-2.00~m/s^2)(4.00~s) + (1.50~m/s^3)(4.00~s)^2$
$v = 12.0~m/s$
When t = 4.00 s, the velocity will be 12.0 m/s
