At time #t = 0#, the position #s# of the particle is
#s(t) = 2(0)^3 -21(0)^2 + 60(0) + 3 = 3#.
The velocity as a function of time is represented by
#v(t) = 6t^2 - 42t + 60#
and the acceleration is
#a(t) = 12t - 42#
(the acceleration is not constant, but its rate of change with time is, since its equation is that of a line)
A graph of position over time:
graph{2x^3 - 21x^2 + 60x + 3 [-10, 20, -10, 100]}
The particle will change direction when its instantaneous velocity is equal to #0#, and we can see from the graph this occurs at times #t = 2# and #t = 5#. If we check this from the velocity equation:
#0 = 6t^2 - 42t + 60#
#0 = t^2 - 7t + 10#
#t = 2, t = 5#
which agrees with the graph.
Here's a graph of the parabolic velocity equation to prove this:
graph{6x^2 - 42x + 60 [-5, 7.5, -20, 40]}
The positions at which the velocity is #0# are
#s(t) = 2(2)^3 -21(2)^2 + 60(2) + 3 = 55#
and
#s(t) = 2(5)^3 -21(5)^2 + 60(5) + 3 = 28#
and its initial velocity is
#v(0) = 6(0)^2 - 42(0) + 60 = 60#
Let's now look at how the velocity changes over time, by examining its acceleration:
#a(t) = 12t - 42#
In the velocity-time graph, if the slope of the tangent line of the parabola at a given time is negative, the acceleration is negative, and if the slope of the tangent line is positive, it's acceleration is positive. Thus, the particle's acceleration is negative (it's slowing down) from time #t = 0# to the time where the velocity-time graph begins to curve upward, which appears to be at time #t = 3.5#. We can find the exact time by finding when the instantaneous acceleration is equal to #0#:
#0 = 12t - 42#
#42 = 12t#
#t = 3.5#
Which is what the graph shows.
In short, from time #t = 0# to #t = 2#, the particle is moving forward at a decreasing speed due to its negative acceleration. At time #t = 2#, when it is 55 units from the origin, the particle begins to move backward with a continually decreasing velocity until time #t = 3.5#, where it begins to gain speed (its velocity is not yet positive though, since it still is moving backward). At time #t = 5#, where it is now 28 units from the origin, its velocity is now positive; it has turned around and is now moving forward with a regularly increasing acceleration, and continues to speed up onward.
