posted by Luna .
A particle moves along the x axis. It is initially at the position 0.350 m, moving with velocity 0.110 m/s and acceleration -0.380 m/s2. Suppose it moves with constant acceleration for 3.50 s.
(a) Find the position of the particle after this time.
(b) Find its velocity at the end of this time interval.
Next, assume it moves with simple harmonic motion for 3.50 s and x = 0 is its equilibrium position.
(c) Find its position.
(d) Find its velocity at the end of this time interval.
Set up the appropriate x(t) or V(t) equation and "plug in" the appropriate t. Thenk crank out the answer. to do, or at least attempt, yourself. Here are the equations:
(a) x = 0.11 t -0.19 t^2
(b) V = 0.11 - 0.38 t
(c) Since t = 3.5 s is the period, it returns to the starting point, x = 0, at that time. You don't need to know the amplitude
(d) Since t = 3.5 s is the period, it returns to the same starting velocity -0.11 m/s
how do you know period is 3.5s