At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . it moves at constant speed. at time t2 = 5.00 s, its acceleration is 4.00 i - 6.00 j m/s2 . what is the radius of the path taken by the particle if t2-t1 is less than one period?

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time , the acceleration vector has direction such that

which indicates the particle is situated at a point on the lower left half of the circle, while at time the acceleration has direction such that

which indicates the particle lies on the upper left half of the circle.

Notice that . That is, the measure of the major arc between the particle's positions at and is 270 degrees, which means that is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

where is the radius of the circle and is the period. We have

and the magnitude of the particle's acceleration toward the center of the circle is

So we find that the path has a radius of

he kicked it 10 feet straight up into the air? i think this works.

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