Answer:
.a vx = -3π
b.vy = 0
c.c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]
d.m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f.f. t = -0.35
g.vx = π - 4πsin (4π(0.124) + π/2)
h.vmax=4π cos (4π(0.045) + π/2)
i.s(t) = [x(t)^2 + y(t)^2]^(1/2)
s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
Explanation:
x(t) = πt + cos(4πt + π/2)
differentiating te orizontal distance wit respect to time t, will give horizontal velocity
change in displacement per change in time is velocity
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
at t =0, substituting te value of t into the above
vx = π - 4π (1)
vx = -3π
b,
y(t) = sin(4πt + π/2)
differentiate wit respect to t
dy/dt=4π cos (4πt + π/2)
π/2=90
when t=0
b. vy=dy/dt = 4π cos (4πt + π/2)
vy = 0
c. slope of te tanent line
y(t)/x(t)
c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]
d. at t=1/6, we substitute into answer gotten in c
m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
at maximum v=0
(4πt + π/2)=0.0043
t=-π/2+0.0043/(4π)
t=-0.124
vx = π - 4πsin (4π(0.124) + π/2)
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
(4πt + π/2=1
t=0.57/4π
t=0.045
vmax=4π cos (4π(0.045) + π/2)
i. resultant of te displacement
s(t) = [x(t)^2 + y(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
s'(t)=
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
