The position vector of a particle as afunction of time is given by :ˆ ˆ( ) 8 cos 3 8 sin 3r t t i t j→= +Determine its velocity and acceleration.Show that both its speed and themagnitude of its acceleration are constant.

To determine the velocity and acceleration of the particle, we need to differentiate the position vector with respect to time.

Given that the position vector is given by r(t) = 8cos(3t)i + 8sin(3t)j, we can differentiate it to find the velocity vector.

Taking the derivative of r(t) with respect to time, we get:

v(t) = d/dt (8cos(3t)i + 8sin(3t)j) = -24sin(3t)i + 24cos(3t)j

So, the velocity vector is v(t) = -24sin(3t)i + 24cos(3t)j.

Next, we can differentiate the velocity vector to find the acceleration vector.

Taking the derivative of v(t) with respect to time, we get:

a(t) = d/dt (-24sin(3t)i + 24cos(3t)j) = -72cos(3t)i - 72sin(3t)j

So, the acceleration vector is a(t) = -72cos(3t)i - 72sin(3t)j.

To show that both the speed and the magnitude of the acceleration are constant, we can calculate them.

The speed of the particle is given by the magnitude of the velocity vector:

|v(t)| = sqrt((-24sin(3t))^2 + (24cos(3t))^2) = sqrt(576sin^2(3t) + 576cos^2(3t)) = sqrt(576(sin^2(3t) + cos^2(3t))) = sqrt(576) = 24

Therefore, the speed of the particle is constant and equal to 24.

Similarly, the magnitude of the acceleration vector is:

|a(t)| = sqrt((-72cos(3t))^2 + (-72sin(3t))^2) = sqrt(5184cos^2(3t) + 5184sin^2(3t)) = sqrt(5184(cos^2(3t) + sin^2(3t))) = sqrt(5184) = 72

Hence, the magnitude of the acceleration is constant and equal to 72.

Therefore, we have shown that both the speed and the magnitude of the acceleration are constant for the given position vector.