A particle is moving in three dimensions and its position vector is given by; overrightarrow (r)(t)=(3,3t^2+2,9t)hat (i)+(2,6t-1,6)hat (j)+(2,7t^3+3,5t)hat (k) where r is in meters and t is in seconds. Determine the magnitude of the instantaneous acceleration at t=3s Express your answer in units of m/s^2 using one decimal place. Yanit: square

To determine the magnitude of the instantaneous acceleration at , we need to find the derivative of the position vector with respect to time and then calculate the magnitude of the resulting acceleration vector.Given the position vector: First, let's find the velocity vector by taking the derivative of the position vector with respect to time: Next, let's find the acceleration vector by taking the derivative of the velocity vector with respect to time: Now, let's substitute into the acceleration vector: Finally, let's calculate the magnitude of the instantaneous acceleration: Therefore, the magnitude of the instantaneous acceleration at is approximately .