<li><p>a particle moves along a horizontal line (positive to the right) according to the law s=t^3 - 3t^2 - 9t +5
during which intervals of time is the particle moving to:
a. the right<br>
b. the left</p></li>
<li><p>a particle moves along a straight line according to the law s = 132 +108t - 16t^2 + 3t^3, s being the distance in meters and t the time in seconds.
a. what is the velocity when t = 2?
b. What is acceleration when t=1 and when t=3?</p></li>
</ol>
<p>bump................</p>
<ol>
<li>You need to compute ds/dt (= the velocity). When this is > 0, the particle is moving to the right, since s would be increasing; when it's < 0, the particle is moving to the left.</li>
</ol>
<p>ds/dt = 3t^2 -6t - 9 = 3(t^2 -2t - 3) </p>
<p>Quick way is to factorize this as 3 (t-3)(t+1) (you could also solve as a quadratic, get the two roots x1 and x2, and rewrite as 3(t-x1) (t-x2) to get the same results).
3 (t-3)(t+1) will be > 0 for very large +ve or -ve t. If you start at a large -ve value for t and then increase t gradually,</p>
<p>ds/dt switches from positive to negative at t = -1
then back to positive at t=3,
and stays positive for all higher values of t.</p>
<ol>
<li>velocity = v(t) = ds/dt = 108 + 32t + 9t^2
acceleration = a(t) = dv/dt = 32 + 18t</li>
</ol>
<p>Find v(2), then a(1) and a(3) for your answers.</p>