The coordinates of an object moving in the xy plane vary with time according to the equations x = -6.85 sin ωt and y = 4.00 - 6.85 cos ωt, where ω is a constant, x and y are in meters, and t is in seconds.
a. Determine the components of the velocity of the object at t = 0.
b. Determine the components of the acceleration of the object at t = 0.
c. Write expressions for the position vector, the velocity vector, and the acceleration vector of the object at any time t > 0.
d. Describe the path of the object in an xy plot.
you say clearly but then Don’t explain why.
a. To determine the components of the velocity of the object at t = 0, we need to differentiate the expressions for x and y with respect to time (t).
Given:
x = -6.85 sin(ωt)
y = 4.00 - 6.85 cos(ωt)
Differentiating x with respect to t:
dx/dt = -6.85 ω cos(ωt)
Differentiating y with respect to t:
dy/dt = 6.85 ω sin(ωt)
At t = 0, we substitute t = 0 into the expressions for velocity to find:
vx = dx/dt = -6.85 ω cos(ω(0)) = -6.85 ω
vy = dy/dt = 6.85 ω sin(ω(0)) = 0
Therefore, at t = 0, the x-component of velocity (vx) is -6.85 ω and the y-component of velocity (vy) is 0.
b. To determine the components of the acceleration of the object at t = 0, we need to differentiate the velocity components with respect to time.
Differentiating vx with respect to t:
d²x/dt² = -6.85 ω² sin(ωt)
Differentiating vy with respect to t:
d²y/dt² = 6.85 ω² cos(ωt)
At t = 0, we substitute t = 0 into the expressions for acceleration to find:
ax = d²x/dt² = -6.85 ω² sin(ω(0)) = 0
ay = d²y/dt² = 6.85 ω² cos(ω(0)) = 6.85 ω²
Therefore, at t = 0, the x-component of acceleration (ax) is 0 and the y-component of acceleration (ay) is 6.85 ω².
c. The position vector (r) of the object at any time t > 0 can be written as:
r = x i + y j
where i and j are unit vectors in the x and y directions, respectively.
The velocity vector (v) of the object at any time t > 0 can be written as:
v = (dx/dt) i + (dy/dt) j
where i and j are the unit vectors in the x and y directions.
And, the acceleration vector (a) of the object at any time t > 0 can be written as:
a = (d²x/dt²) i + (d²y/dt²) j
where i and j are the unit vectors in the x and y directions.
d. The path of the object in an xy plot can be described by the equations x = -6.85 sin(ωt) and y = 4.00 - 6.85 cos(ωt). This represents a circular motion centered at (0, 4.00), with an amplitude of 6.85. The object moves around the center in a clockwise direction, and completes one full revolution in a period of 2π/ω seconds.
To solve this problem, we need to break it down into several steps. Let's go through each part one by one:
a. To determine the components of the velocity at t = 0, we need to differentiate the equations for x and y with respect to time (t). Let's differentiate x first:
dx/dt = d/dt(-6.85 sin ωt)
= -6.85 ω cos ωt
Next, we differentiate y:
dy/dt = d/dt(4.00 - 6.85 cos ωt)
= 6.85 ω sin ωt
Now, we can evaluate the components of velocity at t = 0 by substituting t = 0 into the expressions we just found:
v_x = dx/dt = -6.85 ω cos(ω * 0) = -6.85 ω
v_y = dy/dt = 6.85 ω sin(ω * 0) = 0
So, the components of velocity at t = 0 are v_x = -6.85 ω and v_y = 0.
b. Similarly, to determine the components of acceleration at t = 0, we differentiate the velocity components with respect to t. Differentiating v_x and v_y:
dv_x/dt = d/dt(-6.85 ω) = 0
dv_y/dt = d/dt(0) = 0
Therefore, the components of acceleration at t = 0 are a_x = dv_x/dt = 0 and a_y = dv_y/dt = 0.
c. Now let's find expressions for the position vector r(t), velocity vector v(t), and acceleration vector a(t) at any time t > 0.
Position vector: The position vector in the xy-plane is formed by combining the x and y components: r(t) = x(t)i + y(t)j, where i and j are unit vectors in the x and y directions, respectively:
r(t) = (-6.85 sin ωt)i + (4.00 - 6.85 cos ωt)j
Velocity vector: The velocity vector can be obtained by differentiating the position vector with respect to t:
v(t) = d(r(t))/dt
= d(-6.85 sin ωt)i/dt + d(4.00 - 6.85 cos ωt)j/dt
= -6.85 ω cos ωt)i + 6.85 ω sin ωt)j
Acceleration vector: The acceleration vector can be obtained by differentiating the velocity vector with respect to t:
a(t) = d(v(t))/dt
= d(-6.85 ω cos ωt)i/dt + d(6.85 ω sin ωt)j/dt
= -6.85 ω^2 sin ωt)i + 6.85 ω^2 cos ωt)j
d. The path of the object in an xy plot can be analyzed by looking at the x and y equations separately. The x equation (-6.85 sin ωt) represents harmonic motion with an amplitude of 6.85 and a phase shift of 0. The y equation (4.00 - 6.85 cos ωt) also represents harmonic motion but with an amplitude of 6.85 and a phase shift of π.
Putting these equations together, we can see that the object oscillates along a sinusoidal path with varying x and y coordinates. The exact shape of the path will depend on the value of ω (the constant) and the initial conditions (e.g., the position of the object at t = 0).
In summary, the object moves in a sinusoidal path in the xy-plane, with the x and y coordinates given by x = -6.85 sin ωt and y = 4.00 - 6.85 cos ωt, respectively. The velocity and acceleration components can be obtained by differentiating these equations with respect to time.
dx/dt = -6.85ω cosωt
dy/dt = 6.85ω sinωt
do that again for acceleration, and evaluate at t=0
clearly, the path is a circle with center at (0,4) and radius r=6.85, traced clockwise starting at (0,-2.85)