Note that the total acceleration

  • Thread starter Grover
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  • #1
Here's a problem from my textbook. Hope somebody could give a hand.
The speed of a particle moving in a circle 2m in radius increases at the constant rate of 3m/s^2. At some instant, the magnitude of the total acceleration is 5m/s^2. At this instant, find the centripetal acceleration of the particle and its speed.
 
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  • #2
quantumdude
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Hi, and welcome to PF.

You need to note that the total acceleration is given by:

a=(-v2/r)er+(dv/dt)eφ

where er and eφ are the unit vectors pointing in the radial and tangential directions, respectively.

edit: fixed subscript bracket
 

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