##### Physics problem to solve

label Physics
account_circle Unassigned
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

Oct 3rd, 2015

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

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

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

...
Oct 3rd, 2015
...
Oct 3rd, 2015
Oct 22nd, 2017
check_circle