At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2)i hat + (6.00 m/s2)j. it moves at constant speed. at time t2 = 5.00 s, its acceleration is (6.00 m/s2)i hat + (−7.00 m/s2)j. what is the radius of the path taken by the particle if t2 − t1 is less than one period?

r = 3519.55 m

Explanation:

We know that the acceleration of a particle in a circular motion is directed towards the center of the circumference and has magnitude:

F = rω^2

Where r is the radius of the circumference and ω is the angular velocity.

From the two acceleration vectors we find that their magnitude is

√(7^2+6^2) = √85

Therefore:

√85 m/s^2= rω^2

Now we need to calculate the angular velocity to obtain the radius. Since t2-t1 = 3s is less than one period we can be sure that the angular velocity is equals to the angle traveled between this time divided by 3 s.

The angle with respect to the x-axis  for the particle at t1 and t2 is:

Therefore, the angular velocity ω is (in radians per second):

Therefore:

r = √85 / (0.0511813)^2 = 3519.55 m