Physics problem to solve

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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

Oct 3rd, 2015

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 Vx and Vy be the x and y components of velocity V. Then

Vx = dx/dt

= d{−5.00 sin(ωt)} / dt

=  -5.00ωcos(ωt)

Vy = dy/dt

= d{4.00−5.00 cos(ωt)} / dt  

= 0 – {-5.00ωsin(ωt)}

= 5.00ωsin(ωt)

Acceleration components are ax anday

ax  = dvx/dt

= d{-5.00ωcos(ωt)}/dt

= 5.00ω2sin(ωt)

ay = dvy/dt

= d{5.00ωsin(ωt)}/dt

= 5.00 ω2cos(ωt)

A) components of velocity at t=0

Vx = -5. 00ωcos(ωt)

=  -5.00 ω *cos (0)

= -5.00ω m/s

Vy =  5.00ωsin(ωt)

= 5.00ωsin(0)

=  0 m/s

B) components of acceleration at t=0

ax  = 5.00ω2sin(ωt)

= 5.00ω2sin(0)

= 0 m/s2

ay = 5.00 ω2cos(ωt)

= 5.00ω2cos(0)

= 5.00ω2 m/s2

D) x= −5.00 sin(ωt) ; y =4.00−5.00 cos(ωt)

X2 = -25.00 sin2(ωt)

Y2 = 16.00 – 25.00 cos2(ωt)

X2 + y2 = -25.00 sin2(ωt) +16.00 – 25.00 cos2(ωt)

X2 + y2 = 16.00-25.00 (sin2(ωt) + cos2(ωt) )

X2 + y2 =16.00 -25.00

X2 + y2 = -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.
Sep 20th, 2015

Hi, there is a little correction in the answer.

D) x= −5.00 sin(ωt) ; y =4.00−5.00 cos(ωt)

X2 = 25.00 sin2(ωt)

(Y -4)2 = 25.00 cos2(ωt)

X2 + (y-4)2 = 25.00 sin2(ωt) + 25.00 cos2(ωt)

X2 + (y-4)2 = 25 (sin2(ωt) + cos2(ωt))

X2 + (y-4)2 = 25

X2 + (y-4)2 = 52

Which looks similar to the general equation of a circle is (x-a) 2+ (y-b)2 = r2 so this path is a circle

Sep 20th, 2015

Sorry for the inconvenience caused.

Sep 20th, 2015

All right thank you, actually i got another one!

Sep 20th, 2015

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 unit-vector 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! 

Sep 20th, 2015

Thank you for your understanding

Sep 20th, 2015

sure with pleasure

Sep 20th, 2015

Welcome, i just replied with another question!

Sep 20th, 2015

Thank you i'm wai

Sep 20th, 2015

Waiting ^

Sep 20th, 2015

do you like me to post the answer in the comments are will you be creating a question?


Sep 20th, 2015

Post it here pl

Sep 20th, 2015


Sep 20th, 2015
sp2.png

happy?

Sep 20th, 2015

THANKS ALOT!!


Sep 21st, 2015

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