Question 3 A particle moves on xy plane according to equations: x(t)=2t^3-3t; y(t)=t^2+4 (Take g=10m/s^2 Please mark the closest answer as correct answer ) Find the angle between acceleration and position vectors at t=1 15.5^circ (a) 91.8^circ (b) (c) -6,9^circ (d) 84.9^circ (e) 18.1^circ

To find the angle between acceleration and position vectors, we need to calculate the dot product of the two vectors and then divide it by the product of their magnitudes.<br /><br />First, let's find the acceleration vector by differentiating the position vector with respect to time:<br /><br />$x'(t) = \frac{d}{dt}(2t^3 - 3t) = 6t^2 - 3$<br /><br />$y'(t) = \frac{d}{dt}(t^2 + 4) = 2t$<br /><br />Now, let's find the position vector:<br /><br />$\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} = (2t^3 - 3t)\hat{i} + (t^2 + 4)\hat{j}$<br /><br />Next, let's calculate the dot product of the acceleration and position vectors:<br /><br />$\vec{a} \cdot \vec{r} = (6t^2 - 3)(2t^3 - 3t) + (2t)(t^2 + 4)$<br /><br />Simplifying the expression:<br /><br />$\vec{a} \cdot \vec{r} = 12t^5 - 18t^3 - 6t^3 + 9t + 2t^3 + 8t = 12t^5 - 22t^3 + 17t$<br /><br />Now, let's find the magnitudes of the acceleration and position vectors:<br /><br />$|\vec{a}| = \sqrt{(6t^2 - 3)^2 + (2t)^2} = \sqrt{36t^4 - 36t^2 + 9 + 4t^2} = \sqrt{36t^4 - 32t^2 + 9}$<br /><br />$|\vec{r}| = \sqrt{(2t^3 - 3t)^2 + (t^2 + 4)^2} = \sqrt{4t^6 - 12t^4 + 9t^2 + t^4 + 8t^2 + 16} = \sqrt{4t^6 - 12t^4 + 17t^2 + 16}$<br /><br />Finally, let's calculate the angle between the acceleration and position vectors:<br /><br />$\theta = \arccos\left(\frac{\vec{a} \cdot \vec{r}}{|\vec{a}| |\vec{r}|}\right)$<br /><br />Substituting the values we calculated:<br /><br />$\theta = \arccos\left(\frac{12t^5 - 22t^3 + 17t}{\sqrt{36t^4 - 32t^2 + 9} \cdot \sqrt{4t^6 - 12t^4 + 17t^2 + 16}}\right)$<br /><br />Evaluating this expression at $t=1$:<br /><br />$\theta = \arccos\left(\frac{12 - 22 + 17}{\sqrt{36 - 32 + 9} \cdot \sqrt{4 - 12 + 17 + 16}}\right)$<br /><br />$\theta = \arccos\left(\frac{-13}{\sqrt{13} \cdot \sqrt{27}}\right)$<br /><br />$\theta = \arccos\left(\frac{-13}{\sqrt{13 \cdot 27}}\right)$<br /><br />$\theta = \arccos\left(\frac{-13}{\sqrt{351}}\right)$<br /><br />$\theta = \arccos\left(\frac{-13}{\sqrt{351}}\right)$<br /><br />$\theta = \arccos\left(\frac{-13}{18.73}\right)$<br /><br />$\theta = \arccos\left(-0.695\right)$<br /><br />$\theta = 91.8^{\circ}$<br /><br />Therefore, the correct answer is (a) $91.8^{\circ}$.