A particle is moving in three dimensions and its position vector is given by; overrightarrow (r)(t)=(1,6t^2+2,3t)hat (i)+(1,1t-3,3)hat (j)+(1,7t^3+4,2t)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 find the magnitude of the instantaneous acceleration, we need to find the derivative of the position vector with respect to time and then find the magnitude of the resulting vector.<br /><br />Given the position vector:<br />$\overrightarrow{r}(t) = (1, 6t^2 + 2, 3t)\hat{i} + (1, 1t - 3, 3)\hat{j} + (1, 7t^3 + 4, 2t)\hat{k}$<br /><br />Let's find the derivative of each component with respect to time:<br /><br />$\frac{d}{dt}(1, 6t^2 + 2, 3t)\hat{i} = (0, 12t)\hat{i}$<br />$\frac{d}{dt}(1, 1t - 3, 3)\hat{j} = (0, 1)\hat{j}$<br />$\frac{d}{dt}(1, 7t^3 + 4, 2t)\hat{k} = (0, 21t^2)\hat{k}$<br /><br />Now, let's find the magnitude of the instantaneous acceleration:<br /><br />$|\overrightarrow{a}| = \sqrt{(0, 12t)\hat{i} + (0, 1)\hat{j} + (0, 21t^2)\hat{k}}$<br /><br />Substituting $t = 3$ seconds:<br /><br />$|\overrightarrow{a}| = \sqrt{(0, 36)\hat{i} + (0, 1)\hat{j} + (0, 189)\hat{k}}$<br /><br />$|\overrightarrow{a}| = \sqrt{36 + 1 + 189}$<br /><br />$|\overrightarrow{a}| = \sqrt{226}$<br /><br />$|\overrightarrow{a}| \approx 15.0 \, \text{m/s}^2$<br /><br />Therefore, the magnitude of the instantaneous acceleration at $t = 3$ seconds is approximately 15.0 m/s².