Answer:
a) -2.34 m
b) -1.3 m/s
c) 0.056 m
d) 0.320 m/s
Explanation:
part a
Given:
s(0) = 0.27 m
v(0) = 0.14 m/s
a = -0.320 m/s^2
t = 4.50s
Using kinematic equation of motion for constant acceleration:
s (t) = s(0) + v(0)*t + 0.5*a*t^2
s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2
s(4.5) = -2.34 m
part b
Using kinematic equation of motion for constant acceleration:
v(t) = v(0) + a*t
v(4.5s) = (0.14) + (-0.32)(4.5) = -1.3 m/s
part c
Use equation for simple harmonic motion:
s(t) = A*cos(w*t)
v(t) = -A*w*sin (w*t)
a(t) = -A*(w)^2 * cos (w*t)
0.27 m = A*cos(w*t) .... Eq 1
0.14 m/s = - A*w*sin (w*t) .....Eq2
-0.320 = -A*(w)^2 * cos (w*t) .... Eq3
Solve the three equations above for A, w, and t
Divide 1 and 3:
w^2 = 0.32 / 0.27
w = 1.0887 rad / s
Divide 2 and 1:
w*tan(wt) = 0.14 / 0.27
tan(1.0887*t) = 0.476289
t = 0.4083 s
A = 0.27 / cos (1.0887*0.4083) = 0.3 m
Hence, the SHM is s(t) = 0.3*cos(1.0887*t)
s(4.5) = 0.3*cos(1.0887*4.5) = 0.056 m
part d
v(t) = - 0.32661 * sin (1.0887*t)
v(4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s
