The motion of a particle moving in a circle in the x-y plane is described by the

Question

The motion of a particle moving in a circle in the x-y plane is described by the equations:
r(t) = 5.51

(t) = 5.75t

Where  is the polar angle measured counter-clockwise from the + x-axis in radians, and r is the distance from the origin in m.

A) Calculate the y-coordinate of the particle at the time 2.50 s.
I calculated this with y = r(t)*sin(t)= 5.51*sin(5.57*2.50) = 5.355 m (which was correct)

But I'm having trouble getting the rest of the answers:
B) Calculate the y-component of the velocity at the time 2.20 s?
C)Calculate the magnitude of the acceleration of the particle at the time 3.40 s?
D)By how much does the speed of the particle change from t=10 s to t=71 s?
E)Calculate the x-component of the acceleration at the time 3.40s?

Thank you!

Explanation / Answer

we have
r(t) = 5.51

(t) = 5.75t

A) Calculate the y-coordinate of the particle at the time 2.50 s.

I calculated this with y = r(t)*sin(t)

= 5.51*sin(5.57*2.50) = 5.355 m (which was correct)

But I'm having trouble getting the rest of the answers:

B) Calculate the y-component of the velocity at the time 2.20 s?
vy =   dy/dt = r'(t)sin(t) + r(t)cos(t)'(t)
vy = dy/dt = (0)sin(5.75t) + (5.51)cos(5.75t)(5.75)

vy = dy/dt at 2.2s = 0 + 31.6825 cos (5.75*2.2)

vy = dy/dt at 2.2s = 31.57 m/s

C)Calculate the magnitude of the acceleration of the particle at the time 3.40 s?

x = r cos

x = 5.51 cos(5.75t)

vx = dx/dt = -5.51*5.75 sin(5.75t)

vx = -31.6825 sin(5.75t)

ax = dvx/dt = -31.6825*5.75 cos(5.75t)

ax = -182.17cos(5.75t)

y = r sin

y= 5.51 sin(5.75 t)

vy = dy/dt = 5.51*5.75 cos (5.75t)

v y = 31.6825 cos (5.75t)

ay = dvy/dt = -31.6825*5.75 sin(5.75t)

ay = -182.17 sin(5.75t)

ax at 3.4s = -182.17cos(5.75*3.4) = -139.28
ay at 3.4s = -182.17 sin(5.75*3.4) = - 117.42

a = sqrt(ax^2 +ay^2) = 182.17

D) By how much does the speed of the particle change from t=10 s to t=71 s?

vy at t = 10s is 18.39

vy at t = 71s is 31.29

change in vy is 12.90

vx at t = 10s is -25.80

vx at t = 71s is -4.96

change in vx is -30.75

change in speed is sqrt(vx^2 + vy^2) = 13.43

E) Calculate the x-component of the acceleration at the time 3.40s?

ax at 3.4s = -182.17cos(5.75*3.4) = -139.28  

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