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\question % workshop bank Ch. 5 -> esergel56
Suppose a particle is moving back and forth on a straight line with velocity $v(t) = t^2 - \sqrt{t}$ m/s. We consider the motion of the particle between $t=0$ seconds and $t=4$ seconds.

% v = t^2 - \sqrt{t}
% p = t^3/3 - 2t^{3/2}/3 + C
% a = 2t - 1/2 t^{-1/2}

\begin{parts}

\item What is the particle's displacement during this time period? In other words, at time $t=4$, how far is the particle from the starting point?
% int t^2 - sqrt(t) = t^3/3 - 2t^{3/2}/3 + C |_0^4 = 64/3 - 16/3 = 48/3 = 16 m

\item When is the particle moving forward? When is the particle moving backward?
% The particle is moving forward from t=1 to t=4. It is moving backwards from t=0 to t=1.

\item What is the total distance that the particle traveled during this time period?
% int_1^4 = 16 - 1/3 + 2/3 = 49/3
% int_0^1 = 1/3 - 2/3 - 0 = -1/3
% int_0^4 |v(t)| dt = +1/3 + 49/3 = 50/3

\item When is the particle moving forward the fastest (between $t=0$ and $t=4$)? Moving backward the fastest?
% a = 2t - 1/(2sqrt(t)) = 0 when 4t = sqrt(t) i.e. when t=1/16. Here v(1/16) = 1/16^2 - 1/4 is negative.
% There is no critical point for moving forward. After t=1, the particle is constantly speeding up. So the max positive velocity is at t=4. Here v(4) = 16-2 = 14 m/s

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