At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (5.00 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 + (−5.00 m/s2)j. What is the radius of the path taken by the particle if

t2 − t1
is less than one period

looking at the two accelerations, it appears the need to find the angle.

vi=5+6j
vf=6-5j

costheta=vi dot vj / (magvi * magvj)
maj vi, mag vf=sqrt(25+36)=sqrt61

vi dot vj=30-30=0
cosTheta=0
theta=90


but v=2PI r/12
v^2=4 PI^2 r^2/144

r=v^2/a=pi^2 r^2/sqrt61

r=sqrt61/pi^2
check that carefully. See this http://www.jiskha.com/display.cgi?id=1411303123

To find the radius of the path taken by the particle, we need to first find the net acceleration vector at time t1.

Net acceleration vector at time t1 = (5.00 m/s^2)i hat + (6.00 m/s^2)j

Since the particle is moving at a constant speed, the net acceleration is always directed towards the center of the circular path.

Now, to find the radius of the path, we can use the formula for centripetal acceleration:

a = (v^2) / r

where a is the centripetal acceleration, v is the velocity, and r is the radius.

At time t1, the velocity is perpendicular to the acceleration, so the magnitude of the velocity at t1 is the same as the magnitude of the net acceleration at t1.

|v1| = |net acceleration at t1| = sqrt((5.00 m/s^2)^2 + (6.00 m/s^2)^2) = 7.81 m/s

Now, we can solve for the radius:

r = (v1^2) / a

r = (7.81 m/s)^2 / (5.00 m/s^2)

r = 12.13 m

Therefore, the radius of the path taken by the particle is approximately 12.13 meters.

To find the radius of the path taken by the particle, we can use the formula for the acceleration of an object in circular motion:

a = (v^2) / r

where a is the magnitude of the acceleration, v is the speed of the particle, and r is the radius of the circular path.

Given that the particle moves at a constant speed, the magnitude of the acceleration is the same at both times t1 and t2. Therefore, we can set up the following equations:

(a1)^2 / r1 = (a2)^2 / r2

where a1 and r1 are the acceleration and radius at time t1, and a2 and r2 are the acceleration and radius at time t2.

From the given information:

a1 = (5.00 m/s^2)i + (6.00 m/s^2)j
a2 = (6.00 m/s^2)i + (-5.00 m/s^2)j

To find the speed of the particle, we can calculate the magnitude of the acceleration at either t1 or t2 using the Pythagorean theorem:

|a1| = sqrt((5.00 m/s^2)^2 + (6.00 m/s^2)^2)
|a2| = sqrt((6.00 m/s^2)^2 + (-5.00 m/s^2)^2)

Let's calculate the magnitude of the acceleration at t1:

|a1| = sqrt((5.00 m/s^2)^2 + (6.00 m/s^2)^2)
= sqrt(25.00 + 36.00)
= sqrt(61.00)
≈ 7.81 m/s^2

Similarly, we can calculate the magnitude of the acceleration at t2:

|a2| = sqrt((6.00 m/s^2)^2 + (-5.00 m/s^2)^2)
= sqrt(36.00 + 25.00)
= sqrt(61.00)
≈ 7.81 m/s^2

Since the magnitudes of the accelerations are equal, we can simplify the equation:

|a1|^2 / r1 = |a2|^2 / r2

(7.81 m/s^2)^2 / r1 = (7.81 m/s^2)^2 / r2

By canceling out the common terms:

r2 = r1

Therefore, the radius of the path taken by the particle is the same at both times t1 and t2. Hence, the radius is independent of time.

Note: t2 - t1 being less than one period does not affect the calculation of the radius in this scenario. It only means that the particle has not completed a full revolution within the time interval t2 - t1.