A particle moves along a straight line with the equation of motion s(t) = 1/3t^3 - 2t^2 + 3t where t>0. When does the particle return to it's initial position.
(show work using derivatives i.e the derivative of position-time gives u velocity)
To find when the particle returns to its initial position, we need to find when the position function s(t) is equal to its value at t = 0. First, we find s(0):
s(0) = 1/3(0)^3 - 2(0)^2 + 3(0) = 0
Now, we want to find when s(t) = 0 for t > 0. This is equivalent to finding the roots of s(t).
To do this, we first find the velocity function v(t) by taking the derivative of s(t) with respect to t:
v(t) = ds/dt = d/dt (1/3t^3 - 2t^2 + 3t)
v(t) = (3/3)t^2 - 4t + 3 = t^2 - 4t + 3
Now, we will find the times t when the velocity function is equal to 0, which corresponds to when the position function s(t) is at a maximum or minimum:
v(t) = t^2 - 4t + 3 = 0
This is a quadratic equation, and we can find its roots by factoring:
(t - 1)(t - 3) = 0
So, t = 1 and t = 3 are the critical points when the velocity is 0. We now need to check if the position function s(t) is equal to its initial value (0) at these critical points.
s(1) = 1/3(1)^3 - 2(1)^2 + 3(1) = 1/3 - 2 + 3 = -2/3 + 3 = 7/3 ≠ 0
s(3) = 1/3(3)^3 - 2(3)^2 + 3(3) = 9 - 18 + 9 = 0
Thus, the particle returns to its initial position at t = 3.