##### Physics problem to solve

*label*Physics

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*schedule*1 Day

*account_balance_wallet*$5

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)

**= 5.00****ω**^{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} = 16.00-25.00 (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}
+ (y-4)^{2} = 25.00 sin^{2}(ωt) + 25.00 cos^{2}(ωt)

X^{2}
+ (y-4)^{2} = 25 (sin^{2}(ωt) + cos^{2}(ωt))

X^{2}
+ (y-4)^{2} = 25

X^{2}
+ (y-4)^{2} = 5^{2}

Which
looks similar to the general equation of a circle is (x-a) ^{2}+ (y-b)^{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 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!

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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