- Download PDF
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
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)
The magnitude of the acceleration of the particle is option 2 or 6 m/s^2
We’ve answered 327,574 questions. We can answer yours, too.Ask a question