Answer
(a) At t = 0:
The particle's position is (0,0)
The particle's speed is 2 m/s
At t = 4 s:
The particle's position is (0,0)
The particle's speed is 8.2 m/s
(b) The particle is moving at an angle of $90^{\circ}$ clockwise from the positive x-axis.
The particle is moving at an angle of $14.0^{\circ}$ above the positive x-axis.
Work Step by Step
(a) At t = 0:
$x = (\frac{1}{2}t^3-2t^2)~m$
$x = (\frac{1}{2}(0)^3-2(0)^2)~m$
$x = 0$
$y = (\frac{1}{2}t^2-2t)~m$
$y = (\frac{1}{2}(0)^2-2(0))~m$
$y = 0$
$v_x(t) = \frac{dx}{dt} = (\frac{3}{2}t^2-4t)~m/s$
$v_0 = (\frac{3}{2}(0)^2-4(0))~m/s$
$v_x = 0$
$v_y(t) = \frac{dy}{dt} = (t-2)~m/s$
$v_0 = ((0)-2)~m/s$
$v_y = -2~m/s$
The particle's position is (0,0)
The particle's speed is 2 m/s
At t = 4 s:
$x = (\frac{1}{2}t^3-2t^2)~m$
$x = (\frac{1}{2}(4~s)^3-2(4~s)^2)~m$
$x = 0$
$y = (\frac{1}{2}t^2-2t)~m$
$y = (\frac{1}{2}(4~s)^2-2(4~s))~m$
$y = 0$
$v_x(t) = (\frac{3}{2}t^2-4t)~m/s$
$v_0 = (\frac{3}{2}(4~s)^2-4(4~s))~m/s$
$v_x = 8~m/s$
$v_y(t) = (t-2)~m/s$
$v_0 = ((4~s)-2)~m/s$
$v_y = 2~m/s$
We can find the particle's speed.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(8~m/s)^2+(2~m/s)^2}$
$v = 8.2~m/s$
The particle's position is (0,0)
The particle's speed is 8.2 m/s
(b) At t = 0, the velocity vector points toward the negative y-axis. The particle is moving at an angle of $90^{\circ}$ counterclockwise from the positive x-axis.
We can find the direction of motion at t = 4 seconds as an angle above the positive x-axis.
$tan(\theta) = \frac{v_y}{v_x}$
$\theta = arctan(\frac{2~m/s}{8~m/s})$
$\theta = 14.0^{\circ}$
At t = 4 s, the particle is moving at an angle of $14.0^{\circ}$ above the positive x-axis.
