The coordinates of an object moving in the xy plane vary with time according to

Question

The coordinates of an object moving in the xy plane vary with time according to the equations x =-4.64 sin wt and y = 4.00-4.64 cos wt, where w is a constant, x and y are in meters, and t is in seconds. (a) Determine the components of velocity of the object at t = 0. (Use the following as necessary: w.) vector v=m/s (b) Determine the components of the acceleration of the object at t = 0. (Use the following as necessary: w.) vector a=m/s^2 (c) Write expressions for the position vector, the velocity vector, and the acceleration vector of the object at any time t > 0. (Use the following as necessary: omega for w and t.) vector r= m Vector v= m Vector a= m (d) Describe the path of the object in an xy plot.

Explanation / Answer

Given equations are x = -4.64 sinwt and y = 4.00 - 4.64 coswt

a)
The velocity components of the particle are given by,
vx = dx/dt
vx = d(-4.64sinwt)/dt
vx = -4.64w coswt

vy = dy/dt
vy = d(4.00-4.64coswt)/dt
vy = 4.64w sinwt

At t= 0, vx = -4.64w i^ m/s
             vy = 0
The velocity of the particle is V = -4.64w i^ m/s

b)
The components of the acceleration of the particle
ax = dvx/dt
ax = d(-4.64wcoswt)/dt
ax = 4.64w^2 sinwt
ay = dvy/dt
ay = d(4.64wsinwt)/dt
ay = 4.64w^2 coswt
At t= 0, ax = 0
            ay = 4.64w^2 j^ m/s^2
The acceleration of the particle is, a = 4.64w^2 j^ m/s^2

c)
Position vector, r = (-4.64 sin wt)i^+(4.00 - 4.64 cos wt)j^ m

Velocity vector, v = -4.64w i^ m/s

acceleration vector, a = 4.64w^2 j^ m/s^2

d)
From above considerations,
At t = 0 the particle has only horizontal component of velocity
At t = 0 the particle has only vertical acceleration

This is possible only when the particle is to be in circular motion.
Hence, the particle is moving in a circular path about the origin in the xy plane.

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