Finding Trajectory of x & y in Kinematics

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Thread moved from a technical forum, yada, yada... :-)
I need to find a trajectory

x = 4cos(2t) + 3sin(2t)
y = 3cos(2t) - 4sin(2t)
 
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The trajectory is a function of 2t.
Sine and cosine repeat evert 2π, so the trajectory will repeat. What period?
Plot values of x and y against t.
 
Serbian is easy with google translate...
Page 6 of the .pdf
The question and a solution is presented on that page. I translate, but not all the equations.

Zadatak 2: Kretanje tačke određeno je jednačinama =
Task 2: The motion of a point is determined by equations

𝑥=4cos(2𝑡)+3sin(2𝑡)
𝑦=3cos(2𝑡)−4sin(2𝑡)
(𝑥, 𝑦 - in meters, 𝑡 - in seconds)

Odrediti trajektoriju (putanju), brzinu i ubrzanje tačke u trenutku kada je 𝑡=𝜋[𝑠]
= Determine the trajectory (trajectory), velocity and acceleration of a point at the moment when 𝑡 = 𝜋[𝑠]

Rešenje: = Solution:
Trajektorija (putanja) = Trajectory (path)
Dobijene jednačine kvadrirati i sabrati = Square and add the obtained equations
Brzina = Speed.
Ubrzanje = Acceleration.
 
Baluncore said:
Serbian is easy with google translate...
Page 6 of the .pdf
The question and a solution is presented on that page. I translate, but not all the equations.

Zadatak 2: Kretanje tačke određeno je jednačinama =
Task 2: The motion of a point is determined by equations

𝑥=4cos(2𝑡)+3sin(2𝑡)
𝑦=3cos(2𝑡)−4sin(2𝑡)
(𝑥, 𝑦 - in meters, 𝑡 - in seconds)

Odrediti trajektoriju (putanju), brzinu i ubrzanje tačke u trenutku kada je 𝑡=𝜋[𝑠]
= Determine the trajectory (trajectory), velocity and acceleration of a point at the moment when 𝑡 = 𝜋[𝑠]

Rešenje: = Solution:
Trajektorija (putanja) = Trajectory (path)
Dobijene jednačine kvadrirati i sabrati = Square and add the obtained equations
Brzina = Speed.
Ubrzanje = Acceleration.

I speak Serbian. I understand what is written there, but it was not clear to me why it is done that way.
Now it is.
 
Kitanov said:
I need to find a trajectory

x = 4cos(2t) + 3sin(2t)
y = 3cos(2t) - 4sin(2t)
express ##\cos 2t## and ##\sin 2t## and use ##\cos^2+\sin^2=1##
 
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