Posted by **COFFEE** on Sunday, February 18, 2007 at 7:51pm.

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (9.00i + 8.00j) m/s^2. It moves at constant speed. At time t2 = 3.00s (3/4 of a revolution later), it's acceleration is (8.00i - 9.00j) m/s^2. Find the radius of the path taken by the particle.

Constant speed implies no tangential acceleration. The acceleration is centripetal. So you know it has an angular velocity of displacement/time (solve that, you have the numbers above.)

magnitude acceleration= w^2 * r solve for r.

## Answer This Question

## Related Questions

- Re: Physics/Math - At time t1 = 2.00 s, the acceleration of a particle in ...
- Physics - A particle initially located at the origin has an acceleration of a= 5...
- Physics College - Circular motion - At time t1 = 2.00 s, the acceleration of a ...
- Physics - Two forces act on a 0.200 kg ant: F1 =(3.00i - 8.00j)N and F2=(5.00i...
- Physics - At time t1 = 2.00 s, the acceleration of a particle in ...
- science - m=-2.00i+6.00j-4k and n=2.00i-3.00j. what is the dot product of m and ...
- Physics - a 2.00 kg object has a velocity of 4.00i -3.00j m/s. What is the ...
- physics - Find a vector with a magnitude of 1.00 that bisects the angle between ...
- Physics - A 0.500 kg sphere moving with a velocity (2.00i - 3.50j + 1.00k) m/s ...
- physics - 1.Give examples of a one-dimensional motion where (a) the particle ...

More Related Questions