Physics problem to solve
Physics

Tutor: None Selected  Time limit: 1 Day 
The coordinates of an object moving in the xy plane vary with time according to the equations x= −5.00 sin(ωt)and y =4.00−5.00 cos(ωt) where ω 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 (B Determine the components of 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
Thank you for the opportunity to help you with your question!
Given Data
The position components
x= −5.00 sin(ωt)
y =4.00−5.00 cos(ωt)
Solution
Velocity components are the first derivatives of the position components let V_{x} and V_{y} be the x and y components of velocity V. Then
V_{x} = dx/dt
= d{−5.00 sin(ωt)} / dt
= 5.00ωcos(ωt)
V_{y} = dy/dt
= d{4.00−5.00 cos(ωt)} / dt
= 0 – {5.00ωsin(ωt)}
= 5.00ωsin(ωt)
Acceleration components are a_{x} anda_{y}
a_{x } = dv_{x}/dt
= d{5.00ωcos(ωt)}/dt
= 5.00ω^{2}sin(ωt)
a_{y} = dv_{y}/dt
= d{5.00ωsin(ωt)}/dt
= 5.00 ω^{2}cos(ωt)
A) components of velocity at t=0
V_{x} = 5. 00ωcos(ωt)
= 5.00 ω *cos (0)
= 5.00ω m/s
V_{y} = 5.00ωsin(ωt)
= 5.00ωsin(0)
= 0 m/s
B) components of acceleration at t=0
a_{x } = 5.00ω^{2}sin(ωt)
= 5.00ω^{2}sin(0)
= 0 m/s^{2}
a_{y} = 5.00 ω^{2}cos(ωt)
= 5.00ω^{2}cos(0)
= 5.00ω^{2} m/s^{2}
^{}
D) x= −5.00 sin(ωt) ; y =4.00−5.00 cos(ωt)
X^{2} = 25.00 sin^{2}(ωt)
Y^{2} = 16.00 – 25.00 cos^{2}(ωt)
X^{2} + y^{2} = 25.00 sin^{2}(ωt) +16.00 – 25.00 cos^{2}(ωt)
X^{2} + y^{2} = 16.0025.00 (sin^{2}(ωt) + cos^{2}(ωt) )
X^{2} + y^{2} =16.00 25.00
X^{2} + y^{2} = 9.00
Which is an equation of an circle hence the path of the object is a circle
Please let me know if you need any clarification. I'm always happy to answer your questions.
Hi, there is a little correction in the answer.
D) x= −5.00 sin(ωt) ; y =4.00−5.00 cos(ωt)
X^{2} = 25.00 sin^{2}(ωt)
(Y 4)^{2} = 25.00 cos^{2}(ωt)
X^{2} + (y4)^{2} = 25.00 sin^{2}(ωt) + 25.00 cos^{2}(ωt)
X^{2} + (y4)^{2} = 25 (sin^{2}(ωt) + cos^{2}(ωt))
X^{2} + (y4)^{2} = 25
X^{2} + (y4)^{2} = 5^{2}
Which looks similar to the general equation of a circle is (xa) ^{2}+ (yb)^{2} = r^{2} so this path is a circle
Sorry for the inconvenience caused.
All right thank you, actually i got another one!
A golf ball is hit off a tree at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y = 4.00t − 4.9t 2 , where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball’s position as a function of time, using the unit vectors ˆi and ˆj. By taking derivatives, obtain expression for (b) the velocity vector ~v as a function of time and (c) the acceleration vector ~a as a function of time. (d) Next use unitvector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.00 s.
Could you do this one please!
Welcome, i just replied with another question!
Thank you i'm wai
Waiting ^
do you like me to post the answer in the comments are will you be creating a question?
Post it here pl
THANKS ALOT!!
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