The coordinates of an object moving in the xy plane vary with time according to the equations
x = −6.85 sin ùt and y = 4.00 − 6.85 cos ùt,
where ù 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: ù.)
(b) Determine the components of the acceleration of the object at t = 0. (Use the following as necessary: ù.)
(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 ù and t.)
(d) Describe the path of the object in an xy plot.
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Pre Calculus - Use on
(a) To determine the components of velocity at t = 0, we need to take the derivative of the given equations with respect to time.
The derivative of x with respect to t is obtained by applying the chain rule:
dx/dt = d/dt(-6.85 sin(ωt)) = -6.85ω cos(ωt).
The derivative of y with respect to t is obtained by applying the chain rule:
dy/dt = d/dt(4.00 - 6.85 cos(ωt)) = 6.85ω sin(ωt).
When t = 0, the expressions for the components of velocity become:
Vx = dx/dt = -6.85ω cos(0) = -6.85ω,
Vy = dy/dt = 6.85ω sin(0) = 0.
Therefore, the components of velocity at t = 0 are Vx = -6.85ω and Vy = 0.
(b) Similarly, to determine the components of acceleration at t = 0, we need to take the second derivative of the given equations with respect to time.
The second derivative of x with respect to t is obtained by applying the chain rule again:
d²x/dt² = d/dt(-6.85ω cos(ωt)) = 6.85ω² sin(ωt).
The second derivative of y with respect to t is obtained by applying the chain rule:
d²y/dt² = d/dt(6.85ω sin(ωt)) = 6.85ω² cos(ωt).
When t = 0, the expressions for the components of acceleration become:
Ax = d²x/dt² = 6.85ω² sin(0) = 0,
Ay = d²y/dt² = 6.85ω² cos(0) = 6.85ω².
Therefore, the components of acceleration at t = 0 are Ax = 0 and Ay = 6.85ω².
(c) The position vector of the object at any time t > 0 is given by (x, y):
r(t) = (-6.85 sin(ωt), 4.00 - 6.85 cos(ωt)).
The velocity vector of the object at any time t > 0 is the derivative of the position vector with respect to time:
v(t) = (dx/dt, dy/dt) = (-6.85ω cos(ωt), 6.85ω sin(ωt)).
The acceleration vector of the object at any time t > 0 is the second derivative of the position vector with respect to time:
a(t) = (d²x/dt², d²y/dt²) = (6.85ω² sin(ωt), 6.85ω² cos(ωt)).
(d) The path of the object can be visualized by plotting the x and y coordinates as a function of time on an xy plot. The x-coordinate of the object is given by x = -6.85 sin(ωt), and the y-coordinate is given by y = 4.00 - 6.85 cos(ωt). By varying the value of t, we can calculate the corresponding x and y values and plot them on a graph.
Note: The specifics of the path will depend on the value chosen for ω.
To determine the components of velocity of the object at t = 0, we differentiate the equations for x and y with respect to time.
(a) Velocity components at t = 0:
To determine the velocity components, let's differentiate the equations for x and y with respect to time (t):
For x = -6.85 sin(ωt):
dx/dt = -6.85ωcos(ωt)
For y = 4.00 - 6.85 cos(ωt):
dy/dt = 6.85ωsin(ωt)
At t = 0:
dx/dt = -6.85ωcos(0) = -6.85ω (simplify to -6.85ω, as cos(0) = 1)
dy/dt = 6.85ωsin(0) = 0 (as sin(0) = 0)
So, at t = 0, the x-component of velocity is -6.85ω and the y-component of velocity is 0.
(b) Acceleration components at t = 0:
To determine the acceleration components, we differentiate the velocity components with respect to time (t):
For x = -6.85ω:
d²x/dt² = 0 (since the derivative of a constant is 0)
For y = 0:
d²y/dt² = 0 (since the derivative of a constant is 0)
So, at t = 0, the x-component of acceleration is 0, and the y-component of acceleration is also 0.
(c) Position vector, velocity vector, and acceleration vector:
The position vector of the object at any time t is given by:
r(t) = (x(t), y(t))
r(t) = (-6.85sin(ωt), 4.00 - 6.85cos(ωt))
The velocity vector of the object at any time t is given by the derivatives of the x and y equations with respect to time:
v(t) = (dx/dt, dy/dt)
v(t) = (-6.85ωcos(ωt), 6.85ωsin(ωt))
The acceleration vector of the object at any time t is given by the second derivatives of the x and y equations:
a(t) = (d²x/dt², d²y/dt²)
a(t) = (0, 0)
(d) Path of the object in an xy plot:
To describe the path of the object in an xy plot, we can rearrange the equation for y to get it in terms of x:
y = 4.00 - 6.85cos(ωt)
cos(ωt) = (4.00 - y) / 6.85
Now, substitute this into the equation for x:
x = -6.85sin(ωt) = -6.85sin[ωarccos((4.00 - y)/6.85)]
This equation represents the path of the object in the xy plane. The shape of the path depends on the value of ω, as it determines the frequency and amplitude of the sine and cosine functions.