Originally Posted by

**Grandad** Hello BeckyCI'm guessing from your answer so far that you have made a couple of typos in your question, and it should be

$\displaystyle \vec{r} = (8t-2t^2)\vec{i} + (6 + 4t -t^2)\vec{j}$

In which case, you are right: acceleration $\displaystyle = \vec{\ddot{r}} = -4\vec{i} - 2\vec{j}$

and its magnitude $\displaystyle = |\vec{\ddot{r}}| = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47$

To find the direction, just think about the vector $\displaystyle -4\vec{i} - 2\vec{j}$: this vector is $\displaystyle -4$ units in the $\displaystyle \vec{i}$ direction (i.e. 4 units to the left) and $\displaystyle -2$ units in the $\displaystyle \vec{j}$ direction (i.e. 2 units downward.)

So, draw a right-angled triangle having one side horizontal and 4 units long, and at its left-hand end a side 2 units long vertically downwards. Join the hypotenuse: this side represents the actual acceleration vector. See the attached diagram.

So if the acceleration makes an angle $\displaystyle \theta$ below the horizontal, $\displaystyle \tan\theta = \frac{2}{4}= 0.5 \Rightarrow \theta \approx 26.6^o$.

Grandad