A particle moves along the x axis. It is initially at the position 0.130 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.130 m, moving with velocity 0.190 m/s and acceleration -0.330 m/s2. Suppose it moves with constant acceleration for 3.60 s.

(a) Find the position of the particle after this time (m).
(b) Find its velocity at the end of this time interval (m/s).

Next, assume it moves with simple harmonic motion for 3.60 s and x = 0 is its equilibrium position. (Assume that the velocity and acceleration is the same as in parts (a) and (b).)

(c) Find its position (m).
(d) Find its velocity at the end of this time interval.

Explanation / Answer

a) x = x0 + vx0*t + 1/2*a*t^2 = 0.130 + 0.190*3.60 + 1/2*(-.330)*(3.60)^2 = -1.324m
b) v = vx0 + a*t = 0.190 + (-.330)*3.60 = -0.998m/s
c) Here we have x = A*sin(?t);;so v = ?A*cos(?t) ...and a = -?^2A*sin(?t)
So initially we have x = 0.130 = A*sin(?t); v = 0.190 = ?*A*cos(?t) and a = -0.330 = -?^2A*sin(?t)
So a /x = 0.330/0.130 = ?...so ? = 2.53
Now square the a eqn and the v eqn and add them
x^2 = A^2*sin^2(?t)
v^2 = ?^2A^2*cos^2(?t)
adding we get x^2 + v^2/?^2 = A^2*(sin^2 + cos^2) = A^2
Therefore A = sqrt((x^2 + v^2/?^2)) = sqrt((0.130^2 + 0.190^2/ 2.53^2)) =0.1479 m
So x = 0.1479m*sin(2t) and v = 2*0.1479*cos(2t) = 0.2958*cos(2t)
Note angles are inradians
So at t = 3.60s x = 0.1479*sin(2*3.60) = 0.0185m
and v = 0.2958*cos(2*3.60) = 0.2934m/s

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