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 $t=3s$, we need to find the derivative of the position vector with respect to time and then calculate the magnitude of the resulting acceleration vector.<br /><br />Given the position vector:<br />$\overrightarrow{r}(t) = (3, 3t^2 + 2.9t)\hat{i} + (2.6t - 1.6)\hat{j} + (2.7t^3 + 3.5t)\hat{k}$<br /><br />First, let's find the velocity vector by taking the derivative of the position vector with respect to time:<br />$\overrightarrow{v}(t) = \frac{d\overrightarrow{r}}{dt} = (6t + 2.9)\hat{i} + (2.6)\hat{j} + (8.1t^2 + 3.5)\hat{k}$<br /><br />Next, let's find the acceleration vector by taking the derivative of the velocity vector with respect to time:<br />$\overrightarrow{a}(t) = \frac{d\overrightarrow{v}}{dt} = (6)\hat{i} + (0)\hat{j} + (16.2t)\hat{k}$<br /><br />Now, let's substitute $t=3s$ into the acceleration vector:<br />$\overrightarrow{a}(3) = (6)\hat{i} + (0)\hat{j} + (16.2 \times 3)\hat{k} = (6, 0, 48.6)\hat{k}$<br /><br />Finally, let's calculate the magnitude of the instantaneous acceleration:<br />$|\overrightarrow{a}(3)| = \sqrt{(6)^2 + (0)^2 + (48.6)^2} = \sqrt{36 + 0 + 2373.96} = \sqrt{2409.96} \approx 49.0 \, \text{m/s}^2$<br /><br />Therefore, the magnitude of the instantaneous acceleration at $t=3s$ is approximately $49.0 \, \text{m/s}^2$.