Homework Help

What is the magnitude of the particle's acceleration in the following case:A particle...

user profile pic

pavilion77 | Student, Undergraduate | (Level 2) Honors

Posted February 7, 2012 at 10:54 AM via web

dislike 0 like

What is the magnitude of the particle's acceleration in the following case:

A particle moves in the x-y plane with coordinates given by x=A cos wt  and  y= A sin wt,  where A=1.5 meters and w= 2.0 radians per second .

What is the magnitude of the particle's acceleration?

1. ||a|| = 3.0 m/s^2

2. ||a|| = 6.0 m/s^2

3. ||a|| = 1.3 m/s^2

4. ||a|| = 4.5 m/s^2

5. ||a|| = 0 m/s^2

1 Answer | Add Yours

Top Answer

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 7, 2012 at 7:02 PM (Answer #1)

dislike 2 like

The x-coordinate of the position of the particle is given by A*cos w*t. Here w = 2 radians/second and A = 1.5 m.

The x-coordinate of the position of the particle is: 1.5*cos 2*t

dx/dt = 1.5*2*(-sin 2*t)

d^2x/dt^2 = -6*cos 2t

The y-coordinate of the position of the particle is: 1.5*sin 2*t

dy/dt = 1.5*2*cos 2t

d^2y/dt^2 = -6*sin 2t

The magnitude of the acceleration is sqrt((-6*cos 2t)^2+(-6*sin 2t))

=> 6*sqrt((sin 2t)^2 + (cos 2t)^2)

=> 6

The magnitude of the acceleration of the particle is option 2 or 6 m/s^2

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes